On the ellipse 4x^{2} + 9y^{2} = 1,the points | vedclass

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

Question

(B, D) The given equation of the ellipse is $4x^{2} + 9y^{2} = 1$.Differentiating with respect to $x$,we get $8x + 18yy' = 0$,which implies $y' = -\fr

Vedclass · Accepted Answer

$\left(\frac{2}{5}, -\frac{1}{5}\right)$

Vedclass · Answer

(B, D) The given equation of the ellipse is $4x^{2} + 9y^{2} = 1$.Differentiating with respect to $x$,we get $8x + 18yy' = 0$,which implies $y' = -\frac{8x}{18y} = -\frac{4x}{9y}$.The slope of the line $8x = 9y$ is $y = \frac{8}{9}x$,so the slope $m = \frac{8}{9}$.Since the tangents are parallel to the line,their slopes must be equal: $-\frac{4x}{9y} = \frac{8}{9}$.This simplifies to $-4x = 8y$,or $x = -2y$.Substituting $x = -2y$ into the ellipse equation: $4(-2y)^{2} + 9y^{2} = 1$.$4(4y^{2}) + 9y^{2} = 1$ $\Rightarrow 16y^{2} + 9y^{2} = 1$ $\Rightarrow 25y^{2} = 1$.Thus,$y^{2} = \frac{1}{25}$,which gives $y = \pm \frac{1}{5}$.If $y = \frac{1}{5}$,then $x = -2(\frac{1}{5}) = -\frac{2}{5}$.If $y = -\frac{1}{5}$,then $x = -2(-\frac{1}{5}) = \frac{2}{5}$.The required points are $\left(-\frac{2}{5}, \frac{1}{5}\right)$ and $\left(\frac{2}{5}, -\frac{1}{5}\right)$.

On the ellipse $4x^{2} + 9y^{2} = 1$,the points at which the tangents are parallel to the line $8x = 9y$ are:

Similar Questions

The foci of the ellipse $25x^2 + 4y^2 + 100x - 4y + 100 = 0$ are

$A$ particle is travelling in a clockwise direction on the ellipse $\frac{x^2}{100} + \frac{y^2}{25} = 1$. If the particle leaves the ellipse at the point $(-8, 3)$ and travels along the tangent to the ellipse at that point,then the point where the particle crosses the $Y$-axis is:

The eccentricity of the ellipse $y^{2}+4x^{2}-12x+6y+14=0$ is

The area (in square units) of the quadrilateral formed by joining the foci of the two ellipses $\frac{x^2}{9}+\frac{y^2}{5}=1$ and $\frac{x^2}{5}+\frac{y^2}{9}=1$ is

The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the coordinate axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module