120669
If the product of the lengths of the perpendiculars drawn from the foci to the tangent \(y=\frac{-3}{4} x+3 \sqrt{2}\) of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is 9 , then the eccentricity of that ellipse is
1 \(\frac{\sqrt{2}}{3}\)
2 \(\frac{\sqrt{5}}{6}\)
3 \(\frac{1}{9}\)
4 \(\frac{\sqrt{7}}{4}\)
Explanation:
D We know that,
The product of the length of the perpendicular drawn from the foci to the tangent of the ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { is } \mathrm{b}^2\)
\(\therefore \quad \mathrm{b}^2=9\)
Equation of tangent on line -
\(y=\frac{-3}{4} x+3 \sqrt{2}\)
Here, \(\mathrm{m}=\frac{-3}{4}, \mathrm{c}=3 \sqrt{2}\)
\(\therefore \quad \mathrm{c} = \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
\(3 \sqrt{2} = \pm \sqrt{\mathrm{a}^2\left(\frac{-3}{4}\right)^2+\mathrm{b}^2}\)
\(18 =\frac{9}{16} \mathrm{a}^2+9 \Rightarrow 2-1=\frac{\mathrm{a}^2}{16}\)
\(\mathrm{a}^2 =16\)
\(\mathrm{e} =\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e} =\sqrt{1-\frac{9}{16}}\)
\(\mathrm{e} =\frac{\sqrt{7}}{4}\)
TS EAMCET-10.09.2020
Ellipse
120670
If the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^2}{4}+y^2=1\), then \(c\) is equal to
1 0
2 \(\pm \sqrt{65}\)
3 \(\pm \sqrt{62}\)
4 \(\pm \sqrt{2}\)
5 \(\pm 13\)
Explanation:
B Given,
Equation of straight line
\(y=4 x+c\)
Equation of an ellipse
\(\frac{x^2}{4}+y^2=1\)
From equation (ii)
\(\frac{x^2}{4}+16 x^2+c^2+8 x c=1\)
\(x^2+64 x^2+4 c^2+32 x c=4\)
\(65 x^2+32 c x+4\left(c^2-1\right)=0\)
Since, given line is a tangent to the ellipse.
\(\therefore\) Discriminant \(=0\)
\((32 c)^2-4 \times 65 \times 4\left(c^2-1\right)=0\)
\(1024 c^2-1040\left(c^2-1\right)=0\)
\(1024 c^2-1040 c^2+1040=0\)
\(16 c^2=1040\)
\(c^2=65\)
\(c= \pm \sqrt{65}\)
Kerala CEE-2017
Ellipse
120671
If the tangent to ellipse \(x^2+2 y=1\) at point \(\mathrm{P}\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)\) meets the auxiliary circle at the points \(R\) and \(Q\), then tangents to circle at \(Q\) and \(R\) intersect at
A :
Equation of tangent to ellipse at given point is-
\(\frac{x^2}{1}+\frac{y^2}{1 / 2}=1\)
\(\frac{x_1}{a^2}+\frac{y_1}{b^2}=1\)
\(\frac{x}{\sqrt{2}}+2 y\left(\frac{1}{2}\right)=1\)
\(\frac{x}{\sqrt{2}}+y=1\)
\(x+\sqrt{2} y=\sqrt{2}\)
Now,
QR is chord of contact of circle-
\(x^2+y^2=a^2\)
\(x^2+y^2=1\)
At \(\mathrm{T}(\alpha, \beta)\),
\(\alpha x+\beta y-1=0\)
comparing coefficient of equation (i) and (ii) we have-
\(\alpha=\frac{1}{\sqrt{2}}, \beta=1 \quad \text { so, } \mathrm{T}(\alpha, \beta)=\left(\frac{1}{\sqrt{2}}, 1\right)\)
Manipal UGET-2013
Ellipse
120672
If the line \(l x+m y+n=0\) cuts the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{25}=1\) in points whose eccentoric differ by \(\frac{\pi}{2}\), then \(\frac{a^2 l^2+b^2 m^2}{n^2}\) is equal to
1 1
2 2
3 4
4 \(\frac{3}{2}\)
Explanation:
B Let the points of intersection of the line and the ellipse be \((\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) and
\(\left\{\mathrm{a} \cos \left(\frac{\pi}{2}+\theta\right), \mathrm{b} \sin \left(\frac{\pi}{2}+\theta\right)\right\}=(-\mathrm{a} \sin \theta, \mathrm{b} \cos \theta)\)
since,
they lie on the given line-
\(\mathrm{lx}+\mathrm{my}+\mathrm{n}=0\)
Then,
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta+\mathrm{n}=0\)
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta=-\mathrm{n}\)
\(-\mathrm{la} \sin \theta+\mathrm{mb} \cos \theta+\mathrm{n}=0\)
\(\text { la } \sin \theta-\mathrm{mb} \cos \theta=\mathrm{n}\)
Squaring and adding, we get-
\(\mathrm{l}^2 \mathrm{a}^2 \cos ^2 \theta+\mathrm{m}^2 \mathrm{~b}^2 \theta+2 \mathrm{~m} \text { lab } \sin \theta \cdot \cos \theta+\mathrm{l}^2 \mathrm{a}^2 \sin ^2 \theta+\)
\(\mathrm{m}^2 \mathrm{~b}^2 \cos ^2 \theta-2 \mathrm{~m} l \mathrm{ab} \sin \theta \cos \theta=\mathrm{n}^2+\mathrm{n}^2\)
\(\quad \mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2=2 \mathrm{n}^2\)
\(\Rightarrow \quad \frac{\mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2}{\mathrm{n}^2}=2\)
\(\Rightarrow \quad \text {....(i) }\)
Manipal UGET-2013
Ellipse
120673
The radius of the circle passing through the foci of the ellipse \(\frac{x^2}{4}+\frac{4 y^2}{7}=1\) and having its centre at \(\left(\frac{1}{2}, 2\right)\) is
1 \(\sqrt{5}\)
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given equation of ellipse is \(\frac{x^2}{4}+\frac{y^2}{\frac{7}{4}}=1\)
Here,
\(\mathrm{a}^2=4\)
and
\(\mathrm{b}^2 =\frac{7}{4}\)
\(\mathrm{~b}^2 =\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{7}{4} =4\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}^2 =1-\frac{7}{16}=\frac{9}{16} \Rightarrow \mathrm{e}=\frac{3}{4}\)
Thus, the foci are \(\left( \pm \frac{3}{2}, 0\right)\)
The radius of the required circle having centre \(\left(\frac{1}{2}, 2\right)\) is-
\(=\sqrt{\left(\frac{3}{2}-\frac{1}{2}\right)^2+(0-2)^2}\)
\(=\sqrt{5}\)
120669
If the product of the lengths of the perpendiculars drawn from the foci to the tangent \(y=\frac{-3}{4} x+3 \sqrt{2}\) of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is 9 , then the eccentricity of that ellipse is
1 \(\frac{\sqrt{2}}{3}\)
2 \(\frac{\sqrt{5}}{6}\)
3 \(\frac{1}{9}\)
4 \(\frac{\sqrt{7}}{4}\)
Explanation:
D We know that,
The product of the length of the perpendicular drawn from the foci to the tangent of the ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { is } \mathrm{b}^2\)
\(\therefore \quad \mathrm{b}^2=9\)
Equation of tangent on line -
\(y=\frac{-3}{4} x+3 \sqrt{2}\)
Here, \(\mathrm{m}=\frac{-3}{4}, \mathrm{c}=3 \sqrt{2}\)
\(\therefore \quad \mathrm{c} = \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
\(3 \sqrt{2} = \pm \sqrt{\mathrm{a}^2\left(\frac{-3}{4}\right)^2+\mathrm{b}^2}\)
\(18 =\frac{9}{16} \mathrm{a}^2+9 \Rightarrow 2-1=\frac{\mathrm{a}^2}{16}\)
\(\mathrm{a}^2 =16\)
\(\mathrm{e} =\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e} =\sqrt{1-\frac{9}{16}}\)
\(\mathrm{e} =\frac{\sqrt{7}}{4}\)
TS EAMCET-10.09.2020
Ellipse
120670
If the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^2}{4}+y^2=1\), then \(c\) is equal to
1 0
2 \(\pm \sqrt{65}\)
3 \(\pm \sqrt{62}\)
4 \(\pm \sqrt{2}\)
5 \(\pm 13\)
Explanation:
B Given,
Equation of straight line
\(y=4 x+c\)
Equation of an ellipse
\(\frac{x^2}{4}+y^2=1\)
From equation (ii)
\(\frac{x^2}{4}+16 x^2+c^2+8 x c=1\)
\(x^2+64 x^2+4 c^2+32 x c=4\)
\(65 x^2+32 c x+4\left(c^2-1\right)=0\)
Since, given line is a tangent to the ellipse.
\(\therefore\) Discriminant \(=0\)
\((32 c)^2-4 \times 65 \times 4\left(c^2-1\right)=0\)
\(1024 c^2-1040\left(c^2-1\right)=0\)
\(1024 c^2-1040 c^2+1040=0\)
\(16 c^2=1040\)
\(c^2=65\)
\(c= \pm \sqrt{65}\)
Kerala CEE-2017
Ellipse
120671
If the tangent to ellipse \(x^2+2 y=1\) at point \(\mathrm{P}\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)\) meets the auxiliary circle at the points \(R\) and \(Q\), then tangents to circle at \(Q\) and \(R\) intersect at
A :
Equation of tangent to ellipse at given point is-
\(\frac{x^2}{1}+\frac{y^2}{1 / 2}=1\)
\(\frac{x_1}{a^2}+\frac{y_1}{b^2}=1\)
\(\frac{x}{\sqrt{2}}+2 y\left(\frac{1}{2}\right)=1\)
\(\frac{x}{\sqrt{2}}+y=1\)
\(x+\sqrt{2} y=\sqrt{2}\)
Now,
QR is chord of contact of circle-
\(x^2+y^2=a^2\)
\(x^2+y^2=1\)
At \(\mathrm{T}(\alpha, \beta)\),
\(\alpha x+\beta y-1=0\)
comparing coefficient of equation (i) and (ii) we have-
\(\alpha=\frac{1}{\sqrt{2}}, \beta=1 \quad \text { so, } \mathrm{T}(\alpha, \beta)=\left(\frac{1}{\sqrt{2}}, 1\right)\)
Manipal UGET-2013
Ellipse
120672
If the line \(l x+m y+n=0\) cuts the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{25}=1\) in points whose eccentoric differ by \(\frac{\pi}{2}\), then \(\frac{a^2 l^2+b^2 m^2}{n^2}\) is equal to
1 1
2 2
3 4
4 \(\frac{3}{2}\)
Explanation:
B Let the points of intersection of the line and the ellipse be \((\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) and
\(\left\{\mathrm{a} \cos \left(\frac{\pi}{2}+\theta\right), \mathrm{b} \sin \left(\frac{\pi}{2}+\theta\right)\right\}=(-\mathrm{a} \sin \theta, \mathrm{b} \cos \theta)\)
since,
they lie on the given line-
\(\mathrm{lx}+\mathrm{my}+\mathrm{n}=0\)
Then,
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta+\mathrm{n}=0\)
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta=-\mathrm{n}\)
\(-\mathrm{la} \sin \theta+\mathrm{mb} \cos \theta+\mathrm{n}=0\)
\(\text { la } \sin \theta-\mathrm{mb} \cos \theta=\mathrm{n}\)
Squaring and adding, we get-
\(\mathrm{l}^2 \mathrm{a}^2 \cos ^2 \theta+\mathrm{m}^2 \mathrm{~b}^2 \theta+2 \mathrm{~m} \text { lab } \sin \theta \cdot \cos \theta+\mathrm{l}^2 \mathrm{a}^2 \sin ^2 \theta+\)
\(\mathrm{m}^2 \mathrm{~b}^2 \cos ^2 \theta-2 \mathrm{~m} l \mathrm{ab} \sin \theta \cos \theta=\mathrm{n}^2+\mathrm{n}^2\)
\(\quad \mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2=2 \mathrm{n}^2\)
\(\Rightarrow \quad \frac{\mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2}{\mathrm{n}^2}=2\)
\(\Rightarrow \quad \text {....(i) }\)
Manipal UGET-2013
Ellipse
120673
The radius of the circle passing through the foci of the ellipse \(\frac{x^2}{4}+\frac{4 y^2}{7}=1\) and having its centre at \(\left(\frac{1}{2}, 2\right)\) is
1 \(\sqrt{5}\)
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given equation of ellipse is \(\frac{x^2}{4}+\frac{y^2}{\frac{7}{4}}=1\)
Here,
\(\mathrm{a}^2=4\)
and
\(\mathrm{b}^2 =\frac{7}{4}\)
\(\mathrm{~b}^2 =\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{7}{4} =4\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}^2 =1-\frac{7}{16}=\frac{9}{16} \Rightarrow \mathrm{e}=\frac{3}{4}\)
Thus, the foci are \(\left( \pm \frac{3}{2}, 0\right)\)
The radius of the required circle having centre \(\left(\frac{1}{2}, 2\right)\) is-
\(=\sqrt{\left(\frac{3}{2}-\frac{1}{2}\right)^2+(0-2)^2}\)
\(=\sqrt{5}\)
120669
If the product of the lengths of the perpendiculars drawn from the foci to the tangent \(y=\frac{-3}{4} x+3 \sqrt{2}\) of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is 9 , then the eccentricity of that ellipse is
1 \(\frac{\sqrt{2}}{3}\)
2 \(\frac{\sqrt{5}}{6}\)
3 \(\frac{1}{9}\)
4 \(\frac{\sqrt{7}}{4}\)
Explanation:
D We know that,
The product of the length of the perpendicular drawn from the foci to the tangent of the ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { is } \mathrm{b}^2\)
\(\therefore \quad \mathrm{b}^2=9\)
Equation of tangent on line -
\(y=\frac{-3}{4} x+3 \sqrt{2}\)
Here, \(\mathrm{m}=\frac{-3}{4}, \mathrm{c}=3 \sqrt{2}\)
\(\therefore \quad \mathrm{c} = \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
\(3 \sqrt{2} = \pm \sqrt{\mathrm{a}^2\left(\frac{-3}{4}\right)^2+\mathrm{b}^2}\)
\(18 =\frac{9}{16} \mathrm{a}^2+9 \Rightarrow 2-1=\frac{\mathrm{a}^2}{16}\)
\(\mathrm{a}^2 =16\)
\(\mathrm{e} =\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e} =\sqrt{1-\frac{9}{16}}\)
\(\mathrm{e} =\frac{\sqrt{7}}{4}\)
TS EAMCET-10.09.2020
Ellipse
120670
If the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^2}{4}+y^2=1\), then \(c\) is equal to
1 0
2 \(\pm \sqrt{65}\)
3 \(\pm \sqrt{62}\)
4 \(\pm \sqrt{2}\)
5 \(\pm 13\)
Explanation:
B Given,
Equation of straight line
\(y=4 x+c\)
Equation of an ellipse
\(\frac{x^2}{4}+y^2=1\)
From equation (ii)
\(\frac{x^2}{4}+16 x^2+c^2+8 x c=1\)
\(x^2+64 x^2+4 c^2+32 x c=4\)
\(65 x^2+32 c x+4\left(c^2-1\right)=0\)
Since, given line is a tangent to the ellipse.
\(\therefore\) Discriminant \(=0\)
\((32 c)^2-4 \times 65 \times 4\left(c^2-1\right)=0\)
\(1024 c^2-1040\left(c^2-1\right)=0\)
\(1024 c^2-1040 c^2+1040=0\)
\(16 c^2=1040\)
\(c^2=65\)
\(c= \pm \sqrt{65}\)
Kerala CEE-2017
Ellipse
120671
If the tangent to ellipse \(x^2+2 y=1\) at point \(\mathrm{P}\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)\) meets the auxiliary circle at the points \(R\) and \(Q\), then tangents to circle at \(Q\) and \(R\) intersect at
A :
Equation of tangent to ellipse at given point is-
\(\frac{x^2}{1}+\frac{y^2}{1 / 2}=1\)
\(\frac{x_1}{a^2}+\frac{y_1}{b^2}=1\)
\(\frac{x}{\sqrt{2}}+2 y\left(\frac{1}{2}\right)=1\)
\(\frac{x}{\sqrt{2}}+y=1\)
\(x+\sqrt{2} y=\sqrt{2}\)
Now,
QR is chord of contact of circle-
\(x^2+y^2=a^2\)
\(x^2+y^2=1\)
At \(\mathrm{T}(\alpha, \beta)\),
\(\alpha x+\beta y-1=0\)
comparing coefficient of equation (i) and (ii) we have-
\(\alpha=\frac{1}{\sqrt{2}}, \beta=1 \quad \text { so, } \mathrm{T}(\alpha, \beta)=\left(\frac{1}{\sqrt{2}}, 1\right)\)
Manipal UGET-2013
Ellipse
120672
If the line \(l x+m y+n=0\) cuts the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{25}=1\) in points whose eccentoric differ by \(\frac{\pi}{2}\), then \(\frac{a^2 l^2+b^2 m^2}{n^2}\) is equal to
1 1
2 2
3 4
4 \(\frac{3}{2}\)
Explanation:
B Let the points of intersection of the line and the ellipse be \((\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) and
\(\left\{\mathrm{a} \cos \left(\frac{\pi}{2}+\theta\right), \mathrm{b} \sin \left(\frac{\pi}{2}+\theta\right)\right\}=(-\mathrm{a} \sin \theta, \mathrm{b} \cos \theta)\)
since,
they lie on the given line-
\(\mathrm{lx}+\mathrm{my}+\mathrm{n}=0\)
Then,
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta+\mathrm{n}=0\)
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta=-\mathrm{n}\)
\(-\mathrm{la} \sin \theta+\mathrm{mb} \cos \theta+\mathrm{n}=0\)
\(\text { la } \sin \theta-\mathrm{mb} \cos \theta=\mathrm{n}\)
Squaring and adding, we get-
\(\mathrm{l}^2 \mathrm{a}^2 \cos ^2 \theta+\mathrm{m}^2 \mathrm{~b}^2 \theta+2 \mathrm{~m} \text { lab } \sin \theta \cdot \cos \theta+\mathrm{l}^2 \mathrm{a}^2 \sin ^2 \theta+\)
\(\mathrm{m}^2 \mathrm{~b}^2 \cos ^2 \theta-2 \mathrm{~m} l \mathrm{ab} \sin \theta \cos \theta=\mathrm{n}^2+\mathrm{n}^2\)
\(\quad \mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2=2 \mathrm{n}^2\)
\(\Rightarrow \quad \frac{\mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2}{\mathrm{n}^2}=2\)
\(\Rightarrow \quad \text {....(i) }\)
Manipal UGET-2013
Ellipse
120673
The radius of the circle passing through the foci of the ellipse \(\frac{x^2}{4}+\frac{4 y^2}{7}=1\) and having its centre at \(\left(\frac{1}{2}, 2\right)\) is
1 \(\sqrt{5}\)
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given equation of ellipse is \(\frac{x^2}{4}+\frac{y^2}{\frac{7}{4}}=1\)
Here,
\(\mathrm{a}^2=4\)
and
\(\mathrm{b}^2 =\frac{7}{4}\)
\(\mathrm{~b}^2 =\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{7}{4} =4\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}^2 =1-\frac{7}{16}=\frac{9}{16} \Rightarrow \mathrm{e}=\frac{3}{4}\)
Thus, the foci are \(\left( \pm \frac{3}{2}, 0\right)\)
The radius of the required circle having centre \(\left(\frac{1}{2}, 2\right)\) is-
\(=\sqrt{\left(\frac{3}{2}-\frac{1}{2}\right)^2+(0-2)^2}\)
\(=\sqrt{5}\)
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Ellipse
120669
If the product of the lengths of the perpendiculars drawn from the foci to the tangent \(y=\frac{-3}{4} x+3 \sqrt{2}\) of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is 9 , then the eccentricity of that ellipse is
1 \(\frac{\sqrt{2}}{3}\)
2 \(\frac{\sqrt{5}}{6}\)
3 \(\frac{1}{9}\)
4 \(\frac{\sqrt{7}}{4}\)
Explanation:
D We know that,
The product of the length of the perpendicular drawn from the foci to the tangent of the ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { is } \mathrm{b}^2\)
\(\therefore \quad \mathrm{b}^2=9\)
Equation of tangent on line -
\(y=\frac{-3}{4} x+3 \sqrt{2}\)
Here, \(\mathrm{m}=\frac{-3}{4}, \mathrm{c}=3 \sqrt{2}\)
\(\therefore \quad \mathrm{c} = \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
\(3 \sqrt{2} = \pm \sqrt{\mathrm{a}^2\left(\frac{-3}{4}\right)^2+\mathrm{b}^2}\)
\(18 =\frac{9}{16} \mathrm{a}^2+9 \Rightarrow 2-1=\frac{\mathrm{a}^2}{16}\)
\(\mathrm{a}^2 =16\)
\(\mathrm{e} =\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e} =\sqrt{1-\frac{9}{16}}\)
\(\mathrm{e} =\frac{\sqrt{7}}{4}\)
TS EAMCET-10.09.2020
Ellipse
120670
If the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^2}{4}+y^2=1\), then \(c\) is equal to
1 0
2 \(\pm \sqrt{65}\)
3 \(\pm \sqrt{62}\)
4 \(\pm \sqrt{2}\)
5 \(\pm 13\)
Explanation:
B Given,
Equation of straight line
\(y=4 x+c\)
Equation of an ellipse
\(\frac{x^2}{4}+y^2=1\)
From equation (ii)
\(\frac{x^2}{4}+16 x^2+c^2+8 x c=1\)
\(x^2+64 x^2+4 c^2+32 x c=4\)
\(65 x^2+32 c x+4\left(c^2-1\right)=0\)
Since, given line is a tangent to the ellipse.
\(\therefore\) Discriminant \(=0\)
\((32 c)^2-4 \times 65 \times 4\left(c^2-1\right)=0\)
\(1024 c^2-1040\left(c^2-1\right)=0\)
\(1024 c^2-1040 c^2+1040=0\)
\(16 c^2=1040\)
\(c^2=65\)
\(c= \pm \sqrt{65}\)
Kerala CEE-2017
Ellipse
120671
If the tangent to ellipse \(x^2+2 y=1\) at point \(\mathrm{P}\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)\) meets the auxiliary circle at the points \(R\) and \(Q\), then tangents to circle at \(Q\) and \(R\) intersect at
A :
Equation of tangent to ellipse at given point is-
\(\frac{x^2}{1}+\frac{y^2}{1 / 2}=1\)
\(\frac{x_1}{a^2}+\frac{y_1}{b^2}=1\)
\(\frac{x}{\sqrt{2}}+2 y\left(\frac{1}{2}\right)=1\)
\(\frac{x}{\sqrt{2}}+y=1\)
\(x+\sqrt{2} y=\sqrt{2}\)
Now,
QR is chord of contact of circle-
\(x^2+y^2=a^2\)
\(x^2+y^2=1\)
At \(\mathrm{T}(\alpha, \beta)\),
\(\alpha x+\beta y-1=0\)
comparing coefficient of equation (i) and (ii) we have-
\(\alpha=\frac{1}{\sqrt{2}}, \beta=1 \quad \text { so, } \mathrm{T}(\alpha, \beta)=\left(\frac{1}{\sqrt{2}}, 1\right)\)
Manipal UGET-2013
Ellipse
120672
If the line \(l x+m y+n=0\) cuts the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{25}=1\) in points whose eccentoric differ by \(\frac{\pi}{2}\), then \(\frac{a^2 l^2+b^2 m^2}{n^2}\) is equal to
1 1
2 2
3 4
4 \(\frac{3}{2}\)
Explanation:
B Let the points of intersection of the line and the ellipse be \((\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) and
\(\left\{\mathrm{a} \cos \left(\frac{\pi}{2}+\theta\right), \mathrm{b} \sin \left(\frac{\pi}{2}+\theta\right)\right\}=(-\mathrm{a} \sin \theta, \mathrm{b} \cos \theta)\)
since,
they lie on the given line-
\(\mathrm{lx}+\mathrm{my}+\mathrm{n}=0\)
Then,
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta+\mathrm{n}=0\)
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta=-\mathrm{n}\)
\(-\mathrm{la} \sin \theta+\mathrm{mb} \cos \theta+\mathrm{n}=0\)
\(\text { la } \sin \theta-\mathrm{mb} \cos \theta=\mathrm{n}\)
Squaring and adding, we get-
\(\mathrm{l}^2 \mathrm{a}^2 \cos ^2 \theta+\mathrm{m}^2 \mathrm{~b}^2 \theta+2 \mathrm{~m} \text { lab } \sin \theta \cdot \cos \theta+\mathrm{l}^2 \mathrm{a}^2 \sin ^2 \theta+\)
\(\mathrm{m}^2 \mathrm{~b}^2 \cos ^2 \theta-2 \mathrm{~m} l \mathrm{ab} \sin \theta \cos \theta=\mathrm{n}^2+\mathrm{n}^2\)
\(\quad \mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2=2 \mathrm{n}^2\)
\(\Rightarrow \quad \frac{\mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2}{\mathrm{n}^2}=2\)
\(\Rightarrow \quad \text {....(i) }\)
Manipal UGET-2013
Ellipse
120673
The radius of the circle passing through the foci of the ellipse \(\frac{x^2}{4}+\frac{4 y^2}{7}=1\) and having its centre at \(\left(\frac{1}{2}, 2\right)\) is
1 \(\sqrt{5}\)
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given equation of ellipse is \(\frac{x^2}{4}+\frac{y^2}{\frac{7}{4}}=1\)
Here,
\(\mathrm{a}^2=4\)
and
\(\mathrm{b}^2 =\frac{7}{4}\)
\(\mathrm{~b}^2 =\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{7}{4} =4\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}^2 =1-\frac{7}{16}=\frac{9}{16} \Rightarrow \mathrm{e}=\frac{3}{4}\)
Thus, the foci are \(\left( \pm \frac{3}{2}, 0\right)\)
The radius of the required circle having centre \(\left(\frac{1}{2}, 2\right)\) is-
\(=\sqrt{\left(\frac{3}{2}-\frac{1}{2}\right)^2+(0-2)^2}\)
\(=\sqrt{5}\)
120669
If the product of the lengths of the perpendiculars drawn from the foci to the tangent \(y=\frac{-3}{4} x+3 \sqrt{2}\) of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is 9 , then the eccentricity of that ellipse is
1 \(\frac{\sqrt{2}}{3}\)
2 \(\frac{\sqrt{5}}{6}\)
3 \(\frac{1}{9}\)
4 \(\frac{\sqrt{7}}{4}\)
Explanation:
D We know that,
The product of the length of the perpendicular drawn from the foci to the tangent of the ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { is } \mathrm{b}^2\)
\(\therefore \quad \mathrm{b}^2=9\)
Equation of tangent on line -
\(y=\frac{-3}{4} x+3 \sqrt{2}\)
Here, \(\mathrm{m}=\frac{-3}{4}, \mathrm{c}=3 \sqrt{2}\)
\(\therefore \quad \mathrm{c} = \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}\)
\(3 \sqrt{2} = \pm \sqrt{\mathrm{a}^2\left(\frac{-3}{4}\right)^2+\mathrm{b}^2}\)
\(18 =\frac{9}{16} \mathrm{a}^2+9 \Rightarrow 2-1=\frac{\mathrm{a}^2}{16}\)
\(\mathrm{a}^2 =16\)
\(\mathrm{e} =\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e} =\sqrt{1-\frac{9}{16}}\)
\(\mathrm{e} =\frac{\sqrt{7}}{4}\)
TS EAMCET-10.09.2020
Ellipse
120670
If the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^2}{4}+y^2=1\), then \(c\) is equal to
1 0
2 \(\pm \sqrt{65}\)
3 \(\pm \sqrt{62}\)
4 \(\pm \sqrt{2}\)
5 \(\pm 13\)
Explanation:
B Given,
Equation of straight line
\(y=4 x+c\)
Equation of an ellipse
\(\frac{x^2}{4}+y^2=1\)
From equation (ii)
\(\frac{x^2}{4}+16 x^2+c^2+8 x c=1\)
\(x^2+64 x^2+4 c^2+32 x c=4\)
\(65 x^2+32 c x+4\left(c^2-1\right)=0\)
Since, given line is a tangent to the ellipse.
\(\therefore\) Discriminant \(=0\)
\((32 c)^2-4 \times 65 \times 4\left(c^2-1\right)=0\)
\(1024 c^2-1040\left(c^2-1\right)=0\)
\(1024 c^2-1040 c^2+1040=0\)
\(16 c^2=1040\)
\(c^2=65\)
\(c= \pm \sqrt{65}\)
Kerala CEE-2017
Ellipse
120671
If the tangent to ellipse \(x^2+2 y=1\) at point \(\mathrm{P}\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)\) meets the auxiliary circle at the points \(R\) and \(Q\), then tangents to circle at \(Q\) and \(R\) intersect at
A :
Equation of tangent to ellipse at given point is-
\(\frac{x^2}{1}+\frac{y^2}{1 / 2}=1\)
\(\frac{x_1}{a^2}+\frac{y_1}{b^2}=1\)
\(\frac{x}{\sqrt{2}}+2 y\left(\frac{1}{2}\right)=1\)
\(\frac{x}{\sqrt{2}}+y=1\)
\(x+\sqrt{2} y=\sqrt{2}\)
Now,
QR is chord of contact of circle-
\(x^2+y^2=a^2\)
\(x^2+y^2=1\)
At \(\mathrm{T}(\alpha, \beta)\),
\(\alpha x+\beta y-1=0\)
comparing coefficient of equation (i) and (ii) we have-
\(\alpha=\frac{1}{\sqrt{2}}, \beta=1 \quad \text { so, } \mathrm{T}(\alpha, \beta)=\left(\frac{1}{\sqrt{2}}, 1\right)\)
Manipal UGET-2013
Ellipse
120672
If the line \(l x+m y+n=0\) cuts the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{25}=1\) in points whose eccentoric differ by \(\frac{\pi}{2}\), then \(\frac{a^2 l^2+b^2 m^2}{n^2}\) is equal to
1 1
2 2
3 4
4 \(\frac{3}{2}\)
Explanation:
B Let the points of intersection of the line and the ellipse be \((\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) and
\(\left\{\mathrm{a} \cos \left(\frac{\pi}{2}+\theta\right), \mathrm{b} \sin \left(\frac{\pi}{2}+\theta\right)\right\}=(-\mathrm{a} \sin \theta, \mathrm{b} \cos \theta)\)
since,
they lie on the given line-
\(\mathrm{lx}+\mathrm{my}+\mathrm{n}=0\)
Then,
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta+\mathrm{n}=0\)
\(\text { la } \cos \theta+\mathrm{mb} \sin \theta=-\mathrm{n}\)
\(-\mathrm{la} \sin \theta+\mathrm{mb} \cos \theta+\mathrm{n}=0\)
\(\text { la } \sin \theta-\mathrm{mb} \cos \theta=\mathrm{n}\)
Squaring and adding, we get-
\(\mathrm{l}^2 \mathrm{a}^2 \cos ^2 \theta+\mathrm{m}^2 \mathrm{~b}^2 \theta+2 \mathrm{~m} \text { lab } \sin \theta \cdot \cos \theta+\mathrm{l}^2 \mathrm{a}^2 \sin ^2 \theta+\)
\(\mathrm{m}^2 \mathrm{~b}^2 \cos ^2 \theta-2 \mathrm{~m} l \mathrm{ab} \sin \theta \cos \theta=\mathrm{n}^2+\mathrm{n}^2\)
\(\quad \mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2=2 \mathrm{n}^2\)
\(\Rightarrow \quad \frac{\mathrm{a}^2 \mathrm{l}^2+\mathrm{b}^2 \mathrm{~m}^2}{\mathrm{n}^2}=2\)
\(\Rightarrow \quad \text {....(i) }\)
Manipal UGET-2013
Ellipse
120673
The radius of the circle passing through the foci of the ellipse \(\frac{x^2}{4}+\frac{4 y^2}{7}=1\) and having its centre at \(\left(\frac{1}{2}, 2\right)\) is
1 \(\sqrt{5}\)
2 3
3 \(\sqrt{12}\)
4 \(\frac{7}{2}\)
Explanation:
A Given equation of ellipse is \(\frac{x^2}{4}+\frac{y^2}{\frac{7}{4}}=1\)
Here,
\(\mathrm{a}^2=4\)
and
\(\mathrm{b}^2 =\frac{7}{4}\)
\(\mathrm{~b}^2 =\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(\frac{7}{4} =4\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}^2 =1-\frac{7}{16}=\frac{9}{16} \Rightarrow \mathrm{e}=\frac{3}{4}\)
Thus, the foci are \(\left( \pm \frac{3}{2}, 0\right)\)
The radius of the required circle having centre \(\left(\frac{1}{2}, 2\right)\) is-
\(=\sqrt{\left(\frac{3}{2}-\frac{1}{2}\right)^2+(0-2)^2}\)
\(=\sqrt{5}\)
