120618
The sum of the squares of the eccentricities of the conics \(\frac{x^2}{4}+\frac{y^2}{3}=1\) and \(\frac{x^2}{4}-\frac{y^2}{3}=1\) is
1 2
2 \(\sqrt{\frac{7}{3}}\)
3 \(\sqrt{7}\)
4 \(\sqrt{3}\)
Explanation:
A Given equation of
ellipse \(\frac{x^2}{4}+\frac{y^2}{3}=1\)
So, \(\quad e_1=\sqrt{1-\frac{b^2}{a^2}}\)
\(\mathrm{e}_1=\sqrt{\frac{4-3}{4}}\)
\(\mathrm{e}_1=\frac{1}{2}\)
equation of Hyperbola, \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
So, \(\quad e_2=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}_2=\sqrt{\frac{4+3}{4}}\)
\(\mathrm{e}_2=\frac{\sqrt{7}}{2}\)
Hence, \(\mathrm{e}=\mathrm{e}_1{ }^2+\mathrm{e}_2{ }^2\)
\(\mathrm{e}=\frac{1}{4}+\frac{7}{4}\)
\(\mathrm{e}=\frac{8}{4}\)
\(\mathrm{e}=2\)
Karnataka CET-2013
Ellipse
120619
If \(P\) is point on \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with focii \(S\) and \(S\), then the maximum value of triangle SPS' is
1 ab
2 \(\mathrm{abe}^2\)
3 abe
4 \(a b / e\)
Explanation:
C :
we know that,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(A=\frac{1}{2} \times b \times h\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b} \sin \theta\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b}\)
\(A=\mathrm{aeb}\)
\([\therefore \sin \theta=1]\)
Karnataka CET-2011
Ellipse
120620
If \(x \cos \alpha+y \sin \alpha=4\) is tangent to \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the value of \(\alpha\) is
120621
Area of a triangle formed by tangent and normal to the curve \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) at \(\mathbf{P}\left(\frac{\mathbf{a}}{\sqrt{2}}, \frac{\mathbf{b}}{\sqrt{2}}\right)\) with the \(\mathrm{X}\)-axis is
D :
\(\text { area of } \Delta=\frac{1}{2} \text { Base } \times \text { Height } \left.=\frac{1}{2} \right\rvert\, \text { Normal }|\times| \text { Tangent } \mid\)
\(\text { Length of Normal }=|y| \sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
\(= \frac{b}{\sqrt{2}} \sqrt{1+\left(\frac{-b x}{a \sqrt{a^2-x^2}}\right)_{x=\frac{a}{\sqrt{2}}}^2}\)
\(=\frac{b}{\sqrt{2}} \sqrt{\left(1+\frac{b^2}{a^2}\right)}=\frac{b}{\sqrt{2} a} \sqrt{a^2+b^2}\)
\(\text { Length of tangent }=|y| \sqrt{1+\frac{a^2}{b^2}}=\frac{b}{\sqrt{2} \cdot b} \sqrt{b^2+a^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{2}}\)
\(\text { Area }=\frac{1}{2} \times \frac{b}{\sqrt{2} a} \sqrt{a^2+b^2} \times \frac{\sqrt{a^2+b^2}}{\sqrt{2}} \Rightarrow \frac{b}{4 a}\left(a^2+b^2\right)\)
Karnataka CET-2011
Ellipse
120622
The eccentric angle of the point \((2, \sqrt{3})\) lying on \(\frac{x^2}{16}+\frac{y^2}{4}=1\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
D Given equation of an ellipse is
\(\frac{x^2}{16}+\frac{y^2}{4}=1 \text { and point } P(2, \sqrt{3})\)
Suppose ' \(\theta\) ' be the eccentric angle.
and General equation of an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and
Point \(\mathrm{P}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) on comparing,
\(\mathrm{a}^2=16\) and \(\mathrm{b}^2=4\)
\(\mathrm{a}=4\) and \(\mathrm{b}=2\)
So, point \((4 \cos \theta, 2 \sin \theta\) )
\(4 \cos \theta=2 \text { and } 2 \sin \theta=\sqrt{3}\)
\(\cos \theta=\frac{2}{4} \text { and } \sin \theta=\frac{\sqrt{3}}{2}\)
\(\cos \theta=\cos 60^{\circ} \text { or } \sin \theta=\sin 60^{\circ}\)
\(\theta=60^{\circ} \text { or } \frac{\pi}{3}\)
120618
The sum of the squares of the eccentricities of the conics \(\frac{x^2}{4}+\frac{y^2}{3}=1\) and \(\frac{x^2}{4}-\frac{y^2}{3}=1\) is
1 2
2 \(\sqrt{\frac{7}{3}}\)
3 \(\sqrt{7}\)
4 \(\sqrt{3}\)
Explanation:
A Given equation of
ellipse \(\frac{x^2}{4}+\frac{y^2}{3}=1\)
So, \(\quad e_1=\sqrt{1-\frac{b^2}{a^2}}\)
\(\mathrm{e}_1=\sqrt{\frac{4-3}{4}}\)
\(\mathrm{e}_1=\frac{1}{2}\)
equation of Hyperbola, \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
So, \(\quad e_2=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}_2=\sqrt{\frac{4+3}{4}}\)
\(\mathrm{e}_2=\frac{\sqrt{7}}{2}\)
Hence, \(\mathrm{e}=\mathrm{e}_1{ }^2+\mathrm{e}_2{ }^2\)
\(\mathrm{e}=\frac{1}{4}+\frac{7}{4}\)
\(\mathrm{e}=\frac{8}{4}\)
\(\mathrm{e}=2\)
Karnataka CET-2013
Ellipse
120619
If \(P\) is point on \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with focii \(S\) and \(S\), then the maximum value of triangle SPS' is
1 ab
2 \(\mathrm{abe}^2\)
3 abe
4 \(a b / e\)
Explanation:
C :
we know that,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(A=\frac{1}{2} \times b \times h\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b} \sin \theta\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b}\)
\(A=\mathrm{aeb}\)
\([\therefore \sin \theta=1]\)
Karnataka CET-2011
Ellipse
120620
If \(x \cos \alpha+y \sin \alpha=4\) is tangent to \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the value of \(\alpha\) is
120621
Area of a triangle formed by tangent and normal to the curve \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) at \(\mathbf{P}\left(\frac{\mathbf{a}}{\sqrt{2}}, \frac{\mathbf{b}}{\sqrt{2}}\right)\) with the \(\mathrm{X}\)-axis is
D :
\(\text { area of } \Delta=\frac{1}{2} \text { Base } \times \text { Height } \left.=\frac{1}{2} \right\rvert\, \text { Normal }|\times| \text { Tangent } \mid\)
\(\text { Length of Normal }=|y| \sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
\(= \frac{b}{\sqrt{2}} \sqrt{1+\left(\frac{-b x}{a \sqrt{a^2-x^2}}\right)_{x=\frac{a}{\sqrt{2}}}^2}\)
\(=\frac{b}{\sqrt{2}} \sqrt{\left(1+\frac{b^2}{a^2}\right)}=\frac{b}{\sqrt{2} a} \sqrt{a^2+b^2}\)
\(\text { Length of tangent }=|y| \sqrt{1+\frac{a^2}{b^2}}=\frac{b}{\sqrt{2} \cdot b} \sqrt{b^2+a^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{2}}\)
\(\text { Area }=\frac{1}{2} \times \frac{b}{\sqrt{2} a} \sqrt{a^2+b^2} \times \frac{\sqrt{a^2+b^2}}{\sqrt{2}} \Rightarrow \frac{b}{4 a}\left(a^2+b^2\right)\)
Karnataka CET-2011
Ellipse
120622
The eccentric angle of the point \((2, \sqrt{3})\) lying on \(\frac{x^2}{16}+\frac{y^2}{4}=1\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
D Given equation of an ellipse is
\(\frac{x^2}{16}+\frac{y^2}{4}=1 \text { and point } P(2, \sqrt{3})\)
Suppose ' \(\theta\) ' be the eccentric angle.
and General equation of an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and
Point \(\mathrm{P}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) on comparing,
\(\mathrm{a}^2=16\) and \(\mathrm{b}^2=4\)
\(\mathrm{a}=4\) and \(\mathrm{b}=2\)
So, point \((4 \cos \theta, 2 \sin \theta\) )
\(4 \cos \theta=2 \text { and } 2 \sin \theta=\sqrt{3}\)
\(\cos \theta=\frac{2}{4} \text { and } \sin \theta=\frac{\sqrt{3}}{2}\)
\(\cos \theta=\cos 60^{\circ} \text { or } \sin \theta=\sin 60^{\circ}\)
\(\theta=60^{\circ} \text { or } \frac{\pi}{3}\)
120618
The sum of the squares of the eccentricities of the conics \(\frac{x^2}{4}+\frac{y^2}{3}=1\) and \(\frac{x^2}{4}-\frac{y^2}{3}=1\) is
1 2
2 \(\sqrt{\frac{7}{3}}\)
3 \(\sqrt{7}\)
4 \(\sqrt{3}\)
Explanation:
A Given equation of
ellipse \(\frac{x^2}{4}+\frac{y^2}{3}=1\)
So, \(\quad e_1=\sqrt{1-\frac{b^2}{a^2}}\)
\(\mathrm{e}_1=\sqrt{\frac{4-3}{4}}\)
\(\mathrm{e}_1=\frac{1}{2}\)
equation of Hyperbola, \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
So, \(\quad e_2=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}_2=\sqrt{\frac{4+3}{4}}\)
\(\mathrm{e}_2=\frac{\sqrt{7}}{2}\)
Hence, \(\mathrm{e}=\mathrm{e}_1{ }^2+\mathrm{e}_2{ }^2\)
\(\mathrm{e}=\frac{1}{4}+\frac{7}{4}\)
\(\mathrm{e}=\frac{8}{4}\)
\(\mathrm{e}=2\)
Karnataka CET-2013
Ellipse
120619
If \(P\) is point on \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with focii \(S\) and \(S\), then the maximum value of triangle SPS' is
1 ab
2 \(\mathrm{abe}^2\)
3 abe
4 \(a b / e\)
Explanation:
C :
we know that,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(A=\frac{1}{2} \times b \times h\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b} \sin \theta\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b}\)
\(A=\mathrm{aeb}\)
\([\therefore \sin \theta=1]\)
Karnataka CET-2011
Ellipse
120620
If \(x \cos \alpha+y \sin \alpha=4\) is tangent to \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the value of \(\alpha\) is
120621
Area of a triangle formed by tangent and normal to the curve \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) at \(\mathbf{P}\left(\frac{\mathbf{a}}{\sqrt{2}}, \frac{\mathbf{b}}{\sqrt{2}}\right)\) with the \(\mathrm{X}\)-axis is
D :
\(\text { area of } \Delta=\frac{1}{2} \text { Base } \times \text { Height } \left.=\frac{1}{2} \right\rvert\, \text { Normal }|\times| \text { Tangent } \mid\)
\(\text { Length of Normal }=|y| \sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
\(= \frac{b}{\sqrt{2}} \sqrt{1+\left(\frac{-b x}{a \sqrt{a^2-x^2}}\right)_{x=\frac{a}{\sqrt{2}}}^2}\)
\(=\frac{b}{\sqrt{2}} \sqrt{\left(1+\frac{b^2}{a^2}\right)}=\frac{b}{\sqrt{2} a} \sqrt{a^2+b^2}\)
\(\text { Length of tangent }=|y| \sqrt{1+\frac{a^2}{b^2}}=\frac{b}{\sqrt{2} \cdot b} \sqrt{b^2+a^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{2}}\)
\(\text { Area }=\frac{1}{2} \times \frac{b}{\sqrt{2} a} \sqrt{a^2+b^2} \times \frac{\sqrt{a^2+b^2}}{\sqrt{2}} \Rightarrow \frac{b}{4 a}\left(a^2+b^2\right)\)
Karnataka CET-2011
Ellipse
120622
The eccentric angle of the point \((2, \sqrt{3})\) lying on \(\frac{x^2}{16}+\frac{y^2}{4}=1\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
D Given equation of an ellipse is
\(\frac{x^2}{16}+\frac{y^2}{4}=1 \text { and point } P(2, \sqrt{3})\)
Suppose ' \(\theta\) ' be the eccentric angle.
and General equation of an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and
Point \(\mathrm{P}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) on comparing,
\(\mathrm{a}^2=16\) and \(\mathrm{b}^2=4\)
\(\mathrm{a}=4\) and \(\mathrm{b}=2\)
So, point \((4 \cos \theta, 2 \sin \theta\) )
\(4 \cos \theta=2 \text { and } 2 \sin \theta=\sqrt{3}\)
\(\cos \theta=\frac{2}{4} \text { and } \sin \theta=\frac{\sqrt{3}}{2}\)
\(\cos \theta=\cos 60^{\circ} \text { or } \sin \theta=\sin 60^{\circ}\)
\(\theta=60^{\circ} \text { or } \frac{\pi}{3}\)
120618
The sum of the squares of the eccentricities of the conics \(\frac{x^2}{4}+\frac{y^2}{3}=1\) and \(\frac{x^2}{4}-\frac{y^2}{3}=1\) is
1 2
2 \(\sqrt{\frac{7}{3}}\)
3 \(\sqrt{7}\)
4 \(\sqrt{3}\)
Explanation:
A Given equation of
ellipse \(\frac{x^2}{4}+\frac{y^2}{3}=1\)
So, \(\quad e_1=\sqrt{1-\frac{b^2}{a^2}}\)
\(\mathrm{e}_1=\sqrt{\frac{4-3}{4}}\)
\(\mathrm{e}_1=\frac{1}{2}\)
equation of Hyperbola, \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
So, \(\quad e_2=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}_2=\sqrt{\frac{4+3}{4}}\)
\(\mathrm{e}_2=\frac{\sqrt{7}}{2}\)
Hence, \(\mathrm{e}=\mathrm{e}_1{ }^2+\mathrm{e}_2{ }^2\)
\(\mathrm{e}=\frac{1}{4}+\frac{7}{4}\)
\(\mathrm{e}=\frac{8}{4}\)
\(\mathrm{e}=2\)
Karnataka CET-2013
Ellipse
120619
If \(P\) is point on \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with focii \(S\) and \(S\), then the maximum value of triangle SPS' is
1 ab
2 \(\mathrm{abe}^2\)
3 abe
4 \(a b / e\)
Explanation:
C :
we know that,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(A=\frac{1}{2} \times b \times h\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b} \sin \theta\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b}\)
\(A=\mathrm{aeb}\)
\([\therefore \sin \theta=1]\)
Karnataka CET-2011
Ellipse
120620
If \(x \cos \alpha+y \sin \alpha=4\) is tangent to \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the value of \(\alpha\) is
120621
Area of a triangle formed by tangent and normal to the curve \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) at \(\mathbf{P}\left(\frac{\mathbf{a}}{\sqrt{2}}, \frac{\mathbf{b}}{\sqrt{2}}\right)\) with the \(\mathrm{X}\)-axis is
D :
\(\text { area of } \Delta=\frac{1}{2} \text { Base } \times \text { Height } \left.=\frac{1}{2} \right\rvert\, \text { Normal }|\times| \text { Tangent } \mid\)
\(\text { Length of Normal }=|y| \sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
\(= \frac{b}{\sqrt{2}} \sqrt{1+\left(\frac{-b x}{a \sqrt{a^2-x^2}}\right)_{x=\frac{a}{\sqrt{2}}}^2}\)
\(=\frac{b}{\sqrt{2}} \sqrt{\left(1+\frac{b^2}{a^2}\right)}=\frac{b}{\sqrt{2} a} \sqrt{a^2+b^2}\)
\(\text { Length of tangent }=|y| \sqrt{1+\frac{a^2}{b^2}}=\frac{b}{\sqrt{2} \cdot b} \sqrt{b^2+a^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{2}}\)
\(\text { Area }=\frac{1}{2} \times \frac{b}{\sqrt{2} a} \sqrt{a^2+b^2} \times \frac{\sqrt{a^2+b^2}}{\sqrt{2}} \Rightarrow \frac{b}{4 a}\left(a^2+b^2\right)\)
Karnataka CET-2011
Ellipse
120622
The eccentric angle of the point \((2, \sqrt{3})\) lying on \(\frac{x^2}{16}+\frac{y^2}{4}=1\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
D Given equation of an ellipse is
\(\frac{x^2}{16}+\frac{y^2}{4}=1 \text { and point } P(2, \sqrt{3})\)
Suppose ' \(\theta\) ' be the eccentric angle.
and General equation of an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and
Point \(\mathrm{P}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) on comparing,
\(\mathrm{a}^2=16\) and \(\mathrm{b}^2=4\)
\(\mathrm{a}=4\) and \(\mathrm{b}=2\)
So, point \((4 \cos \theta, 2 \sin \theta\) )
\(4 \cos \theta=2 \text { and } 2 \sin \theta=\sqrt{3}\)
\(\cos \theta=\frac{2}{4} \text { and } \sin \theta=\frac{\sqrt{3}}{2}\)
\(\cos \theta=\cos 60^{\circ} \text { or } \sin \theta=\sin 60^{\circ}\)
\(\theta=60^{\circ} \text { or } \frac{\pi}{3}\)
120618
The sum of the squares of the eccentricities of the conics \(\frac{x^2}{4}+\frac{y^2}{3}=1\) and \(\frac{x^2}{4}-\frac{y^2}{3}=1\) is
1 2
2 \(\sqrt{\frac{7}{3}}\)
3 \(\sqrt{7}\)
4 \(\sqrt{3}\)
Explanation:
A Given equation of
ellipse \(\frac{x^2}{4}+\frac{y^2}{3}=1\)
So, \(\quad e_1=\sqrt{1-\frac{b^2}{a^2}}\)
\(\mathrm{e}_1=\sqrt{\frac{4-3}{4}}\)
\(\mathrm{e}_1=\frac{1}{2}\)
equation of Hyperbola, \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
So, \(\quad e_2=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}_2=\sqrt{\frac{4+3}{4}}\)
\(\mathrm{e}_2=\frac{\sqrt{7}}{2}\)
Hence, \(\mathrm{e}=\mathrm{e}_1{ }^2+\mathrm{e}_2{ }^2\)
\(\mathrm{e}=\frac{1}{4}+\frac{7}{4}\)
\(\mathrm{e}=\frac{8}{4}\)
\(\mathrm{e}=2\)
Karnataka CET-2013
Ellipse
120619
If \(P\) is point on \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with focii \(S\) and \(S\), then the maximum value of triangle SPS' is
1 ab
2 \(\mathrm{abe}^2\)
3 abe
4 \(a b / e\)
Explanation:
C :
we know that,
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(A=\frac{1}{2} \times b \times h\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b} \sin \theta\)
\(A=\frac{1}{2} \times 2 \mathrm{ae} \times \mathrm{b}\)
\(A=\mathrm{aeb}\)
\([\therefore \sin \theta=1]\)
Karnataka CET-2011
Ellipse
120620
If \(x \cos \alpha+y \sin \alpha=4\) is tangent to \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the value of \(\alpha\) is
120621
Area of a triangle formed by tangent and normal to the curve \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) at \(\mathbf{P}\left(\frac{\mathbf{a}}{\sqrt{2}}, \frac{\mathbf{b}}{\sqrt{2}}\right)\) with the \(\mathrm{X}\)-axis is
D :
\(\text { area of } \Delta=\frac{1}{2} \text { Base } \times \text { Height } \left.=\frac{1}{2} \right\rvert\, \text { Normal }|\times| \text { Tangent } \mid\)
\(\text { Length of Normal }=|y| \sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
\(= \frac{b}{\sqrt{2}} \sqrt{1+\left(\frac{-b x}{a \sqrt{a^2-x^2}}\right)_{x=\frac{a}{\sqrt{2}}}^2}\)
\(=\frac{b}{\sqrt{2}} \sqrt{\left(1+\frac{b^2}{a^2}\right)}=\frac{b}{\sqrt{2} a} \sqrt{a^2+b^2}\)
\(\text { Length of tangent }=|y| \sqrt{1+\frac{a^2}{b^2}}=\frac{b}{\sqrt{2} \cdot b} \sqrt{b^2+a^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{2}}\)
\(\text { Area }=\frac{1}{2} \times \frac{b}{\sqrt{2} a} \sqrt{a^2+b^2} \times \frac{\sqrt{a^2+b^2}}{\sqrt{2}} \Rightarrow \frac{b}{4 a}\left(a^2+b^2\right)\)
Karnataka CET-2011
Ellipse
120622
The eccentric angle of the point \((2, \sqrt{3})\) lying on \(\frac{x^2}{16}+\frac{y^2}{4}=1\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
D Given equation of an ellipse is
\(\frac{x^2}{16}+\frac{y^2}{4}=1 \text { and point } P(2, \sqrt{3})\)
Suppose ' \(\theta\) ' be the eccentric angle.
and General equation of an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and
Point \(\mathrm{P}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)\) on comparing,
\(\mathrm{a}^2=16\) and \(\mathrm{b}^2=4\)
\(\mathrm{a}=4\) and \(\mathrm{b}=2\)
So, point \((4 \cos \theta, 2 \sin \theta\) )
\(4 \cos \theta=2 \text { and } 2 \sin \theta=\sqrt{3}\)
\(\cos \theta=\frac{2}{4} \text { and } \sin \theta=\frac{\sqrt{3}}{2}\)
\(\cos \theta=\cos 60^{\circ} \text { or } \sin \theta=\sin 60^{\circ}\)
\(\theta=60^{\circ} \text { or } \frac{\pi}{3}\)
