120665
The normal drawn at the point \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\) to the ellipse \(\frac{x^2}{9}+\frac{y^2}{7}=1\) intersects its major axis at the point
1 \(\left(0, \sqrt{\frac{2}{7}}\right)\)
2 \(\left(-\sqrt{\frac{2}{9}}, 0\right)\)
3 \(\left(0,-\sqrt{\frac{2}{7}}\right)\)
4 \(\left(\sqrt{\frac{2}{9}}, 0\right)\)
Explanation:
D Given, the ellipse \(\frac{x^2}{9}+\frac{y^2}{7}=1\)
Differentiating w.r.t. \(x\), we get -
\(\frac{2 x}{9}+\frac{2 y}{7} \frac{d y}{d x}=0\)
\(\therefore \frac{d y}{d x}=\frac{-2 x / 9}{2 y / 7}=\frac{-x}{y}\left(\frac{7}{9}\right)\)
Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at point \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\)
\(=\frac{-\sqrt{9} \cos \frac{\pi}{4}}{\sqrt{7} \sin \frac{\pi}{4}} \cdot \frac{7}{9}=-\sqrt{\frac{7}{9}}\)
\(\therefore\) Slope of the normal, \(\mathrm{m}=\frac{3}{\sqrt{7}}\)
\(\therefore\) Equation of this normal \(\mathrm{y}=\frac{3}{\sqrt{7}} \mathrm{x}+\mathrm{c}\)
But this line passes through \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\)
\(\therefore \sqrt{7} \cdot \sin \frac{\pi}{4}=\frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \cos \frac{\pi}{4}+\mathrm{c}\)
\(\therefore \mathrm{c}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{3 \times 3}{\sqrt{2} \cdot \sqrt{7}}\)
\(\quad=\frac{1}{\sqrt{2}}\left[\frac{7-9}{\sqrt{7}}\right]=\frac{-2}{\sqrt{14}}\)
\(\therefore\) Equation of the normal is \(\mathrm{y}=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
Since this line intersects its major axis
\(\therefore \mathrm{y}=0\)
\(\therefore 0=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
\(\mathrm{x}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3}=\frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3}=\frac{\sqrt{2}}{3}=\sqrt{\frac{2}{9}}\)
\(\therefore \mathrm{y}=0\)
\(\therefore 0=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
\(\mathrm{x}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3}=\frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3}=\frac{\sqrt{2}}{3}=\sqrt{\frac{2}{9}}\)
\(\therefore\) The point of intersection is \(\left(\sqrt{\frac{2}{9}}, 0\right)\).
TS EAMCET-04.05.2019
Ellipse
120666
The locus of the mid-points of the portion of the tangents of the ellipse \(\frac{x^2}{2}+\frac{y^2}{1}=1\) intercepted between the coordinate axes is
C Given, equation of ellipse \(\frac{\mathrm{x}^2}{2}+\frac{\mathrm{y}^2}{1}=1\)
Here, \(\mathrm{a}^2=2, \mathrm{~b}^2=1\)
Equaiton of tangent of ellipse are
\(\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1\)
The \(x\)-intercept and y-intercept are \((\sqrt{2} \sec \theta, 0)\) and \((0\), \(\operatorname{cosec} \theta\) ) respectively.
Let, (h, k) mid-point of intercept -
\(\mathrm{h}=\frac{\sqrt{2} \sec \theta}{2}, \quad \mathrm{k}=\frac{\operatorname{cosec} \theta}{2}\)
\(\cos \theta=\frac{1}{\sqrt{2} \mathrm{~h}} \text { and } \sin \theta=\frac{1}{2 \mathrm{k}}\)
Squaring and adding, we get -
\(\frac{1}{2 \mathrm{~h}^2}+\frac{1}{4 \mathrm{k}^2}=1\)
\(\therefore \text { locus be } \frac{1}{2 \mathrm{x}^2}+\frac{1}{4 \mathrm{y}^2}=1\)
TS EAMCET-03.05.2019
Ellipse
120667
If a circle \((x-1)^2+y^2=r^2\) touches the ellipse \(x^2\) \(+4 y^2=16\) internally, then \(r=\)
1 \(\sqrt{\frac{11}{3}}\)
2 \(\frac{11}{3}\)
3 \(\sqrt{\frac{15}{2}}\)
4 2
Explanation:
A Given, the circle \((x-1)^2+y^2=r^2\)
Centre of the circle is \((1,0)\)
Given the equation of the ellipse \(=x^2+4 y^2=16\)
Point of the ellipse \(=(4 \cos \theta, 2 \sin \theta)\)
Equation of normal of ellipse
\(\Rightarrow\) ax \(\sec \theta-\) by \(\operatorname{cosec} \theta=\mathrm{a}^2-\mathrm{b}^2\)
\(\Rightarrow 4 \mathrm{x} \sec \theta-2 \mathrm{y} \operatorname{cosec} \theta=12\)
It passes through \((1,0)\)
\(\Rightarrow \sec \theta=3\)
\(\Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3} \cos \theta=\frac{1}{3}\)
Now, point will be \(\left(\frac{4}{3}, \frac{4 \sqrt{2}}{3}\right)\)
\(\text { Radius } =\text { Distance of }(1,0) \text { and }\left(\frac{4}{3}, \frac{4 \sqrt{2}}{3}\right)\)
\(=\sqrt{\left(1-\frac{4}{3}\right)^2+\left(0-\frac{4 \sqrt{2}}{3}\right)^2}\)
\(=\sqrt{\frac{1}{9}+\frac{32}{9}}=\sqrt{\frac{33}{9}}=\sqrt{\frac{11}{3}}\)
TS EAMCET-05.08.2021
Ellipse
120668
The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse \(S=\frac{x^2}{16}+\frac{y^2}{12}=1\) is
1 96
2 16
3 128
4 64
Explanation:
D Given ellipse, \(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{12}=1\)
End point of latus rectum \(=\left( \pm \mathrm{ae}, \frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)=(2,3)\)
Equation of tangent at \((2,3)\) of ellipse \(\frac{x^2}{16}+\frac{y^2}{12}=1\) is,
\(\frac{2 \mathrm{x}}{16}+\frac{3 \mathrm{y}}{12}=1 \Rightarrow \frac{\mathrm{x}}{8}+\frac{\mathrm{y}}{4}=1\)
Area of quadrilateral \(\mathrm{ABCD}=4 \times \frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\)
\(=4 \times \frac{1}{2} \times 8 \times 4=64\)
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Ellipse
120665
The normal drawn at the point \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\) to the ellipse \(\frac{x^2}{9}+\frac{y^2}{7}=1\) intersects its major axis at the point
1 \(\left(0, \sqrt{\frac{2}{7}}\right)\)
2 \(\left(-\sqrt{\frac{2}{9}}, 0\right)\)
3 \(\left(0,-\sqrt{\frac{2}{7}}\right)\)
4 \(\left(\sqrt{\frac{2}{9}}, 0\right)\)
Explanation:
D Given, the ellipse \(\frac{x^2}{9}+\frac{y^2}{7}=1\)
Differentiating w.r.t. \(x\), we get -
\(\frac{2 x}{9}+\frac{2 y}{7} \frac{d y}{d x}=0\)
\(\therefore \frac{d y}{d x}=\frac{-2 x / 9}{2 y / 7}=\frac{-x}{y}\left(\frac{7}{9}\right)\)
Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at point \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\)
\(=\frac{-\sqrt{9} \cos \frac{\pi}{4}}{\sqrt{7} \sin \frac{\pi}{4}} \cdot \frac{7}{9}=-\sqrt{\frac{7}{9}}\)
\(\therefore\) Slope of the normal, \(\mathrm{m}=\frac{3}{\sqrt{7}}\)
\(\therefore\) Equation of this normal \(\mathrm{y}=\frac{3}{\sqrt{7}} \mathrm{x}+\mathrm{c}\)
But this line passes through \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\)
\(\therefore \sqrt{7} \cdot \sin \frac{\pi}{4}=\frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \cos \frac{\pi}{4}+\mathrm{c}\)
\(\therefore \mathrm{c}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{3 \times 3}{\sqrt{2} \cdot \sqrt{7}}\)
\(\quad=\frac{1}{\sqrt{2}}\left[\frac{7-9}{\sqrt{7}}\right]=\frac{-2}{\sqrt{14}}\)
\(\therefore\) Equation of the normal is \(\mathrm{y}=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
Since this line intersects its major axis
\(\therefore \mathrm{y}=0\)
\(\therefore 0=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
\(\mathrm{x}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3}=\frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3}=\frac{\sqrt{2}}{3}=\sqrt{\frac{2}{9}}\)
\(\therefore \mathrm{y}=0\)
\(\therefore 0=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
\(\mathrm{x}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3}=\frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3}=\frac{\sqrt{2}}{3}=\sqrt{\frac{2}{9}}\)
\(\therefore\) The point of intersection is \(\left(\sqrt{\frac{2}{9}}, 0\right)\).
TS EAMCET-04.05.2019
Ellipse
120666
The locus of the mid-points of the portion of the tangents of the ellipse \(\frac{x^2}{2}+\frac{y^2}{1}=1\) intercepted between the coordinate axes is
C Given, equation of ellipse \(\frac{\mathrm{x}^2}{2}+\frac{\mathrm{y}^2}{1}=1\)
Here, \(\mathrm{a}^2=2, \mathrm{~b}^2=1\)
Equaiton of tangent of ellipse are
\(\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1\)
The \(x\)-intercept and y-intercept are \((\sqrt{2} \sec \theta, 0)\) and \((0\), \(\operatorname{cosec} \theta\) ) respectively.
Let, (h, k) mid-point of intercept -
\(\mathrm{h}=\frac{\sqrt{2} \sec \theta}{2}, \quad \mathrm{k}=\frac{\operatorname{cosec} \theta}{2}\)
\(\cos \theta=\frac{1}{\sqrt{2} \mathrm{~h}} \text { and } \sin \theta=\frac{1}{2 \mathrm{k}}\)
Squaring and adding, we get -
\(\frac{1}{2 \mathrm{~h}^2}+\frac{1}{4 \mathrm{k}^2}=1\)
\(\therefore \text { locus be } \frac{1}{2 \mathrm{x}^2}+\frac{1}{4 \mathrm{y}^2}=1\)
TS EAMCET-03.05.2019
Ellipse
120667
If a circle \((x-1)^2+y^2=r^2\) touches the ellipse \(x^2\) \(+4 y^2=16\) internally, then \(r=\)
1 \(\sqrt{\frac{11}{3}}\)
2 \(\frac{11}{3}\)
3 \(\sqrt{\frac{15}{2}}\)
4 2
Explanation:
A Given, the circle \((x-1)^2+y^2=r^2\)
Centre of the circle is \((1,0)\)
Given the equation of the ellipse \(=x^2+4 y^2=16\)
Point of the ellipse \(=(4 \cos \theta, 2 \sin \theta)\)
Equation of normal of ellipse
\(\Rightarrow\) ax \(\sec \theta-\) by \(\operatorname{cosec} \theta=\mathrm{a}^2-\mathrm{b}^2\)
\(\Rightarrow 4 \mathrm{x} \sec \theta-2 \mathrm{y} \operatorname{cosec} \theta=12\)
It passes through \((1,0)\)
\(\Rightarrow \sec \theta=3\)
\(\Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3} \cos \theta=\frac{1}{3}\)
Now, point will be \(\left(\frac{4}{3}, \frac{4 \sqrt{2}}{3}\right)\)
\(\text { Radius } =\text { Distance of }(1,0) \text { and }\left(\frac{4}{3}, \frac{4 \sqrt{2}}{3}\right)\)
\(=\sqrt{\left(1-\frac{4}{3}\right)^2+\left(0-\frac{4 \sqrt{2}}{3}\right)^2}\)
\(=\sqrt{\frac{1}{9}+\frac{32}{9}}=\sqrt{\frac{33}{9}}=\sqrt{\frac{11}{3}}\)
TS EAMCET-05.08.2021
Ellipse
120668
The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse \(S=\frac{x^2}{16}+\frac{y^2}{12}=1\) is
1 96
2 16
3 128
4 64
Explanation:
D Given ellipse, \(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{12}=1\)
End point of latus rectum \(=\left( \pm \mathrm{ae}, \frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)=(2,3)\)
Equation of tangent at \((2,3)\) of ellipse \(\frac{x^2}{16}+\frac{y^2}{12}=1\) is,
\(\frac{2 \mathrm{x}}{16}+\frac{3 \mathrm{y}}{12}=1 \Rightarrow \frac{\mathrm{x}}{8}+\frac{\mathrm{y}}{4}=1\)
Area of quadrilateral \(\mathrm{ABCD}=4 \times \frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\)
\(=4 \times \frac{1}{2} \times 8 \times 4=64\)
120665
The normal drawn at the point \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\) to the ellipse \(\frac{x^2}{9}+\frac{y^2}{7}=1\) intersects its major axis at the point
1 \(\left(0, \sqrt{\frac{2}{7}}\right)\)
2 \(\left(-\sqrt{\frac{2}{9}}, 0\right)\)
3 \(\left(0,-\sqrt{\frac{2}{7}}\right)\)
4 \(\left(\sqrt{\frac{2}{9}}, 0\right)\)
Explanation:
D Given, the ellipse \(\frac{x^2}{9}+\frac{y^2}{7}=1\)
Differentiating w.r.t. \(x\), we get -
\(\frac{2 x}{9}+\frac{2 y}{7} \frac{d y}{d x}=0\)
\(\therefore \frac{d y}{d x}=\frac{-2 x / 9}{2 y / 7}=\frac{-x}{y}\left(\frac{7}{9}\right)\)
Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at point \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\)
\(=\frac{-\sqrt{9} \cos \frac{\pi}{4}}{\sqrt{7} \sin \frac{\pi}{4}} \cdot \frac{7}{9}=-\sqrt{\frac{7}{9}}\)
\(\therefore\) Slope of the normal, \(\mathrm{m}=\frac{3}{\sqrt{7}}\)
\(\therefore\) Equation of this normal \(\mathrm{y}=\frac{3}{\sqrt{7}} \mathrm{x}+\mathrm{c}\)
But this line passes through \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\)
\(\therefore \sqrt{7} \cdot \sin \frac{\pi}{4}=\frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \cos \frac{\pi}{4}+\mathrm{c}\)
\(\therefore \mathrm{c}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{3 \times 3}{\sqrt{2} \cdot \sqrt{7}}\)
\(\quad=\frac{1}{\sqrt{2}}\left[\frac{7-9}{\sqrt{7}}\right]=\frac{-2}{\sqrt{14}}\)
\(\therefore\) Equation of the normal is \(\mathrm{y}=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
Since this line intersects its major axis
\(\therefore \mathrm{y}=0\)
\(\therefore 0=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
\(\mathrm{x}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3}=\frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3}=\frac{\sqrt{2}}{3}=\sqrt{\frac{2}{9}}\)
\(\therefore \mathrm{y}=0\)
\(\therefore 0=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
\(\mathrm{x}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3}=\frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3}=\frac{\sqrt{2}}{3}=\sqrt{\frac{2}{9}}\)
\(\therefore\) The point of intersection is \(\left(\sqrt{\frac{2}{9}}, 0\right)\).
TS EAMCET-04.05.2019
Ellipse
120666
The locus of the mid-points of the portion of the tangents of the ellipse \(\frac{x^2}{2}+\frac{y^2}{1}=1\) intercepted between the coordinate axes is
C Given, equation of ellipse \(\frac{\mathrm{x}^2}{2}+\frac{\mathrm{y}^2}{1}=1\)
Here, \(\mathrm{a}^2=2, \mathrm{~b}^2=1\)
Equaiton of tangent of ellipse are
\(\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1\)
The \(x\)-intercept and y-intercept are \((\sqrt{2} \sec \theta, 0)\) and \((0\), \(\operatorname{cosec} \theta\) ) respectively.
Let, (h, k) mid-point of intercept -
\(\mathrm{h}=\frac{\sqrt{2} \sec \theta}{2}, \quad \mathrm{k}=\frac{\operatorname{cosec} \theta}{2}\)
\(\cos \theta=\frac{1}{\sqrt{2} \mathrm{~h}} \text { and } \sin \theta=\frac{1}{2 \mathrm{k}}\)
Squaring and adding, we get -
\(\frac{1}{2 \mathrm{~h}^2}+\frac{1}{4 \mathrm{k}^2}=1\)
\(\therefore \text { locus be } \frac{1}{2 \mathrm{x}^2}+\frac{1}{4 \mathrm{y}^2}=1\)
TS EAMCET-03.05.2019
Ellipse
120667
If a circle \((x-1)^2+y^2=r^2\) touches the ellipse \(x^2\) \(+4 y^2=16\) internally, then \(r=\)
1 \(\sqrt{\frac{11}{3}}\)
2 \(\frac{11}{3}\)
3 \(\sqrt{\frac{15}{2}}\)
4 2
Explanation:
A Given, the circle \((x-1)^2+y^2=r^2\)
Centre of the circle is \((1,0)\)
Given the equation of the ellipse \(=x^2+4 y^2=16\)
Point of the ellipse \(=(4 \cos \theta, 2 \sin \theta)\)
Equation of normal of ellipse
\(\Rightarrow\) ax \(\sec \theta-\) by \(\operatorname{cosec} \theta=\mathrm{a}^2-\mathrm{b}^2\)
\(\Rightarrow 4 \mathrm{x} \sec \theta-2 \mathrm{y} \operatorname{cosec} \theta=12\)
It passes through \((1,0)\)
\(\Rightarrow \sec \theta=3\)
\(\Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3} \cos \theta=\frac{1}{3}\)
Now, point will be \(\left(\frac{4}{3}, \frac{4 \sqrt{2}}{3}\right)\)
\(\text { Radius } =\text { Distance of }(1,0) \text { and }\left(\frac{4}{3}, \frac{4 \sqrt{2}}{3}\right)\)
\(=\sqrt{\left(1-\frac{4}{3}\right)^2+\left(0-\frac{4 \sqrt{2}}{3}\right)^2}\)
\(=\sqrt{\frac{1}{9}+\frac{32}{9}}=\sqrt{\frac{33}{9}}=\sqrt{\frac{11}{3}}\)
TS EAMCET-05.08.2021
Ellipse
120668
The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse \(S=\frac{x^2}{16}+\frac{y^2}{12}=1\) is
1 96
2 16
3 128
4 64
Explanation:
D Given ellipse, \(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{12}=1\)
End point of latus rectum \(=\left( \pm \mathrm{ae}, \frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)=(2,3)\)
Equation of tangent at \((2,3)\) of ellipse \(\frac{x^2}{16}+\frac{y^2}{12}=1\) is,
\(\frac{2 \mathrm{x}}{16}+\frac{3 \mathrm{y}}{12}=1 \Rightarrow \frac{\mathrm{x}}{8}+\frac{\mathrm{y}}{4}=1\)
Area of quadrilateral \(\mathrm{ABCD}=4 \times \frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\)
\(=4 \times \frac{1}{2} \times 8 \times 4=64\)
120665
The normal drawn at the point \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\) to the ellipse \(\frac{x^2}{9}+\frac{y^2}{7}=1\) intersects its major axis at the point
1 \(\left(0, \sqrt{\frac{2}{7}}\right)\)
2 \(\left(-\sqrt{\frac{2}{9}}, 0\right)\)
3 \(\left(0,-\sqrt{\frac{2}{7}}\right)\)
4 \(\left(\sqrt{\frac{2}{9}}, 0\right)\)
Explanation:
D Given, the ellipse \(\frac{x^2}{9}+\frac{y^2}{7}=1\)
Differentiating w.r.t. \(x\), we get -
\(\frac{2 x}{9}+\frac{2 y}{7} \frac{d y}{d x}=0\)
\(\therefore \frac{d y}{d x}=\frac{-2 x / 9}{2 y / 7}=\frac{-x}{y}\left(\frac{7}{9}\right)\)
Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at point \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\)
\(=\frac{-\sqrt{9} \cos \frac{\pi}{4}}{\sqrt{7} \sin \frac{\pi}{4}} \cdot \frac{7}{9}=-\sqrt{\frac{7}{9}}\)
\(\therefore\) Slope of the normal, \(\mathrm{m}=\frac{3}{\sqrt{7}}\)
\(\therefore\) Equation of this normal \(\mathrm{y}=\frac{3}{\sqrt{7}} \mathrm{x}+\mathrm{c}\)
But this line passes through \(\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)\)
\(\therefore \sqrt{7} \cdot \sin \frac{\pi}{4}=\frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \cos \frac{\pi}{4}+\mathrm{c}\)
\(\therefore \mathrm{c}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{3 \times 3}{\sqrt{2} \cdot \sqrt{7}}\)
\(\quad=\frac{1}{\sqrt{2}}\left[\frac{7-9}{\sqrt{7}}\right]=\frac{-2}{\sqrt{14}}\)
\(\therefore\) Equation of the normal is \(\mathrm{y}=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
Since this line intersects its major axis
\(\therefore \mathrm{y}=0\)
\(\therefore 0=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
\(\mathrm{x}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3}=\frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3}=\frac{\sqrt{2}}{3}=\sqrt{\frac{2}{9}}\)
\(\therefore \mathrm{y}=0\)
\(\therefore 0=\frac{3}{\sqrt{7}} \mathrm{x}-\frac{2}{\sqrt{14}}\)
\(\mathrm{x}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3}=\frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3}=\frac{\sqrt{2}}{3}=\sqrt{\frac{2}{9}}\)
\(\therefore\) The point of intersection is \(\left(\sqrt{\frac{2}{9}}, 0\right)\).
TS EAMCET-04.05.2019
Ellipse
120666
The locus of the mid-points of the portion of the tangents of the ellipse \(\frac{x^2}{2}+\frac{y^2}{1}=1\) intercepted between the coordinate axes is
C Given, equation of ellipse \(\frac{\mathrm{x}^2}{2}+\frac{\mathrm{y}^2}{1}=1\)
Here, \(\mathrm{a}^2=2, \mathrm{~b}^2=1\)
Equaiton of tangent of ellipse are
\(\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1\)
The \(x\)-intercept and y-intercept are \((\sqrt{2} \sec \theta, 0)\) and \((0\), \(\operatorname{cosec} \theta\) ) respectively.
Let, (h, k) mid-point of intercept -
\(\mathrm{h}=\frac{\sqrt{2} \sec \theta}{2}, \quad \mathrm{k}=\frac{\operatorname{cosec} \theta}{2}\)
\(\cos \theta=\frac{1}{\sqrt{2} \mathrm{~h}} \text { and } \sin \theta=\frac{1}{2 \mathrm{k}}\)
Squaring and adding, we get -
\(\frac{1}{2 \mathrm{~h}^2}+\frac{1}{4 \mathrm{k}^2}=1\)
\(\therefore \text { locus be } \frac{1}{2 \mathrm{x}^2}+\frac{1}{4 \mathrm{y}^2}=1\)
TS EAMCET-03.05.2019
Ellipse
120667
If a circle \((x-1)^2+y^2=r^2\) touches the ellipse \(x^2\) \(+4 y^2=16\) internally, then \(r=\)
1 \(\sqrt{\frac{11}{3}}\)
2 \(\frac{11}{3}\)
3 \(\sqrt{\frac{15}{2}}\)
4 2
Explanation:
A Given, the circle \((x-1)^2+y^2=r^2\)
Centre of the circle is \((1,0)\)
Given the equation of the ellipse \(=x^2+4 y^2=16\)
Point of the ellipse \(=(4 \cos \theta, 2 \sin \theta)\)
Equation of normal of ellipse
\(\Rightarrow\) ax \(\sec \theta-\) by \(\operatorname{cosec} \theta=\mathrm{a}^2-\mathrm{b}^2\)
\(\Rightarrow 4 \mathrm{x} \sec \theta-2 \mathrm{y} \operatorname{cosec} \theta=12\)
It passes through \((1,0)\)
\(\Rightarrow \sec \theta=3\)
\(\Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3} \cos \theta=\frac{1}{3}\)
Now, point will be \(\left(\frac{4}{3}, \frac{4 \sqrt{2}}{3}\right)\)
\(\text { Radius } =\text { Distance of }(1,0) \text { and }\left(\frac{4}{3}, \frac{4 \sqrt{2}}{3}\right)\)
\(=\sqrt{\left(1-\frac{4}{3}\right)^2+\left(0-\frac{4 \sqrt{2}}{3}\right)^2}\)
\(=\sqrt{\frac{1}{9}+\frac{32}{9}}=\sqrt{\frac{33}{9}}=\sqrt{\frac{11}{3}}\)
TS EAMCET-05.08.2021
Ellipse
120668
The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse \(S=\frac{x^2}{16}+\frac{y^2}{12}=1\) is
1 96
2 16
3 128
4 64
Explanation:
D Given ellipse, \(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{12}=1\)
End point of latus rectum \(=\left( \pm \mathrm{ae}, \frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)=(2,3)\)
Equation of tangent at \((2,3)\) of ellipse \(\frac{x^2}{16}+\frac{y^2}{12}=1\) is,
\(\frac{2 \mathrm{x}}{16}+\frac{3 \mathrm{y}}{12}=1 \Rightarrow \frac{\mathrm{x}}{8}+\frac{\mathrm{y}}{4}=1\)
Area of quadrilateral \(\mathrm{ABCD}=4 \times \frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\)
\(=4 \times \frac{1}{2} \times 8 \times 4=64\)
