A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$
A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$
- [IIT 2013]
- A
$6$
- B
$7$
- C
$8$
- D
$9$
Similar Questions
Consider the ellipse
$\frac{x^2}{4}+\frac{y^2}{3}=1$
Let $H (\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
$List-I$
$List-II$
If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is
($P$) $\frac{(\sqrt{3}-1)^4}{8}$
If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is
($Q$) $1$
If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is
($R$) $\frac{3}{4}$
If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is
($S$) $\frac{1}{2 \sqrt{3}}$
($T$) $\frac{3 \sqrt{3}}{2}$
The correct option is:
Consider the ellipse
$\frac{x^2}{4}+\frac{y^2}{3}=1$
Let $H (\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
| $List-I$ | $List-II$ |
| If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | ($P$) $\frac{(\sqrt{3}-1)^4}{8}$ |
| If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | ($Q$) $1$ |
| If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | ($R$) $\frac{3}{4}$ |
| If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | ($S$) $\frac{1}{2 \sqrt{3}}$ |
| ($T$) $\frac{3 \sqrt{3}}{2}$ |
The correct option is:
- [IIT 2022]
If the point of intersections of the ellipse $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b ^{2}}=1$ and the circle $x ^{2}+ y ^{2}=4 b , b > 4$ lie on the curve $y^{2}=3 x^{2},$ then $b$ is equal to:
If the point of intersections of the ellipse $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b ^{2}}=1$ and the circle $x ^{2}+ y ^{2}=4 b , b > 4$ lie on the curve $y^{2}=3 x^{2},$ then $b$ is equal to:
- [JEE MAIN 2021]
For the ellipse $3{x^2} + 4{y^2} = 12$, the length of latus rectum is
For the ellipse $3{x^2} + 4{y^2} = 12$, the length of latus rectum is
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The angle of intersection of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and circle ${x^2} + {y^2} = ab$, is
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b < a)$, be a ellipse with major axis $A B$ and minor axis $C D$. Let $F_1$ and $F_2$ be its two foci, with $A, F_1, F_2, B$ in that order on the segment $A B$. Suppose $\angle F_1 C B=90^{\circ}$. The eccentricity of the ellipse is
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b < a)$, be a ellipse with major axis $A B$ and minor axis $C D$. Let $F_1$ and $F_2$ be its two foci, with $A, F_1, F_2, B$ in that order on the segment $A B$. Suppose $\angle F_1 C B=90^{\circ}$. The eccentricity of the ellipse is
- [KVPY 2020]