Length of chord of ellipse frac{x^2}{25}+frac{y^2}{16}=1 , whose mid point is left(1, frac{2}{5}righ

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10-2. Parabola, Ellipse, Hyperbola

hard

The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to:

$\frac{\sqrt{1691}}{5}$

$\frac{\sqrt{2009}}{5}$

$\frac{\sqrt{1741}}{5}$

$\frac{\sqrt{1541}}{5}$

(JEE MAIN-2024)

Solution

Equation of chord with given middle point.

$\mathrm{T}=\mathrm{S}_1$

$ \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} $

$ \frac{8 x+5 y}{200}=\frac{8+2}{200}$

$y=\frac{10-8 x}{5}$ $…….(i)$

$\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1$ (put in original equation)

$\frac{16 \mathrm{x}^2+100+64 \mathrm{x}^2-160 \mathrm{x}}{400}=1$

$ 4 x^2-8 x-15=0 $

$ x=\frac{8 \pm \sqrt{304}}{8} $

$x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8}$

$\mathrm{x}_1=\frac{8+\sqrt{304}}{8} ; \mathrm{x}_2=\frac{8-\sqrt{304}}{8}$

Similarly, $y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}$

$\mathrm{y}_1=\frac{2-\sqrt{304}}{5} ; \mathrm{y}_2=\frac{2+\sqrt{304}}{5}$

Distance =$\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}$

$=\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5}$

Standard 11

Mathematics

The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi /4$ is :

normal

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:

normal

(IIT-2022)

The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9 x^2+16 y^2=144$ is

normal

(KVPY-2011)

If the normal at the point $P(\theta )$ to the ellipse $\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1$ intersects it again at the point $Q(2\theta )$, then $\cos \theta $ is equal to

hard

In a triangle $A B C$ with fixed base $B C$, the vertex $A$ moves such that $\cos B+\cos C=4 \sin ^2 \frac{A}{2} .$ If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$, respectively, then

$(A)$ $b+c=4 a$

$(B)$ $b+c=2 a$

$(C)$ locus of point $A$ is an ellipse

$(D)$ locus of point $A$ is a pair of straight lines

normal

(IIT-2009)

$List-I$	$List-II$
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 length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to:

Solution

Similar Questions

The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi /4$ is :

The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9 x^2+16 y^2=144$ is

If the normal at the point $P(\theta )$ to the ellipse $\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1$ intersects it again at the point $Q(2\theta )$, then $\cos \theta $ is equal to

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