If P_1 and P_2 are two points on the ellipse | vedclass

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

Question

(D) The slope of the chord joining $(0, 1)$ and $(2, 0)$ is $m = \frac{0 - 1}{2 - 0} = -\frac{1}{2}$.Since the tangents at $P_1$ and $P_2$ are paralle

Vedclass · Accepted Answer

$\sqrt{10}$

Vedclass · Answer

(D) The slope of the chord joining $(0, 1)$ and $(2, 0)$ is $m = \frac{0 - 1}{2 - 0} = -\frac{1}{2}$.Since the tangents at $P_1$ and $P_2$ are parallel to this chord,their slope is $m = -\frac{1}{2}$.The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Here $a^2 = 4, b^2 = 1$,and $m = -\frac{1}{2}$.Thus,$y = -\frac{1}{2}x \pm \sqrt{4(-\frac{1}{2})^2 + 1} = -\frac{1}{2}x \pm \sqrt{1 + 1} = -\frac{1}{2}x \pm \sqrt{2}$.Substituting $y = -\frac{1}{2}x + \sqrt{2}$ into the ellipse equation $\frac{x^2}{4} + y^2 = 1$:$\frac{x^2}{4} + (-\frac{x}{2} + \sqrt{2})^2 = 1$ $\Rightarrow \frac{x^2}{4} + \frac{x^2}{4} - \sqrt{2}x + 2 = 1$ $\Rightarrow \frac{x^2}{2} - \sqrt{2}x + 1 = 0$ $\Rightarrow x^2 - 2\sqrt{2}x + 2 = 0$ $\Rightarrow (x - \sqrt{2})^2 = 0$.So,$x = \sqrt{2}$,which gives $y = -\frac{1}{2}(\sqrt{2}) + \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Point $P_1 = (\sqrt{2}, \frac{1}{\sqrt{2}})$.Similarly,for $y = -\frac{1}{2}x - \sqrt{2}$,we get $x = -\sqrt{2}$ and $y = -\frac{1}{\sqrt{2}}$. Point $P_2 = (-\sqrt{2}, -\frac{1}{\sqrt{2}})$.The distance $P_1P_2 = \sqrt{(\sqrt{2} - (-\sqrt{2}))^2 + (\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}))^2} = \sqrt{(2\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{8 + 2} = \sqrt{10}$.

If $P_1$ and $P_2$ are two points on the ellipse $\frac{x^2}{4} + y^2 = 1$ at which the tangents are parallel to the chord joining the points $(0, 1)$ and $(2, 0)$,then the distance between $P_1$ and $P_2$ is

Similar Questions

The area of the region bounded by the curve $9x^2 + 4y^2 - 36 = 0$ is (in $\pi$)

The coordinates of the foci of the ellipse $3x^2 + 4y^2 - 12x - 8y + 4 = 0$ are

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ is inscribed in a rectangle of dimensions $2a$ and $2b$ respectively. If the angle between the diagonals of the rectangle is $\tan^{-1}(4\sqrt{3})$,then the eccentricity of that ellipse is

$A$ point moves such that the sum of its distances from $(ae, 0)$ and $(-ae, 0)$ is $2a$. Then the equation to its locus,where $b^2 = a^2(1 - e^2)$,is

The centre of the ellipse $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module