Let for two distinct values of p the lines y= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The condition for the line $y = mx + p$ to touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $p^2 = a^2m^2 + b^2$. Here $a^2 = 16, b^2

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) The condition for the line $y = mx + p$ to touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $p^2 = a^2m^2 + b^2$. Here $a^2 = 16, b^2 = 9, m = 1$,so $p^2 = 16(1)^2 + 9 = 25$,which gives $p = \pm 5$.The points of contact are given by $\left( \mp \frac{a^2m}{p}, \pm \frac{b^2}{p} ight)$. For $p = 5$,$A = \left( -\frac{16}{5}, \frac{9}{5} ight)$. For $p = -5$,$B = \left( \frac{16}{5}, -\frac{9}{5} ight)$.The line $y = x$ intersects the ellipse $\frac{x^2}{16} + \frac{x^2}{9} = 1$,so $x^2(\frac{9+16}{144}) = 1$,$x^2 = \frac{144}{25}$,$x = \pm \frac{12}{5}$. Thus $C = \left( \frac{12}{5}, \frac{12}{5} ight)$ and $D = \left( -\frac{12}{5}, -\frac{12}{5} ight)$.The quadrilateral $ABCD$ is a parallelogram because the lines $y = x + 5$ and $y = x - 5$ are parallel,and the line $y = x$ passes through the center $(0,0)$.The area of the quadrilateral $ABCD$ can be calculated as $2 	imes 	ext{Area}(	riangle ABC)$.Using the coordinates $A(-\frac{16}{5}, \frac{9}{5})$,$B(\frac{16}{5}, -\frac{9}{5})$,$C(\frac{12}{5}, \frac{12}{5})$,the area is $2 	imes \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = |-\frac{16}{5}(-\frac{9}{5} - \frac{12}{5}) + \frac{16}{5}(\frac{12}{5} - \frac{9}{5}) + \frac{12}{5}(\frac{9}{5} - (-\frac{9}{5}))| = |-\frac{16}{5}(-\frac{21}{5}) + \frac{16}{5}(\frac{3}{5}) + \frac{12}{5}(\frac{18}{5})| = |\frac{336}{25} + \frac{48}{25} + \frac{216}{25}| = \frac{600}{25} = 24$.

Let for two distinct values of $p$ the lines $y=x+p$ touch the ellipse $E: \frac{x^2}{16} + \frac{y^2}{9} = 1$ at the points $A$ and $B$. Let the line $y = x$ intersect $E$ at the points $C$ and $D$. Then the area of the quadrilateral $ABCD$ is equal to

Similar Questions

Find the area of the triangle formed by the $X$-axis and the tangent and the normal to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$.

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes is

Let $S = 0$ be an ellipse whose vertices are the extremities of the minor axis of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. If $S = 0$ passes through the foci of $E$,then its eccentricity is (considering the eccentricity of $E$ as $e$).

Equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are:

The length of the latus rectum of the ellipse $5x^2 + 9y^2 = 45$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module