If the curves y^2=6x and 9x^2+by^2=16 interse | vedclass

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

Question

(A) Given curves are $y^2=6x$ $(i)$ and $9x^2+by^2=16$ $(ii)$.For $(i)$,differentiating with respect to $x$: $2y \frac{dy}{dx} = 6 \Rightarrow \frac{d

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(A) Given curves are $y^2=6x$ $(i)$ and $9x^2+by^2=16$ $(ii)$.For $(i)$,differentiating with respect to $x$: $2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y}$.For $(ii)$,differentiating with respect to $x$: $18x + 2by \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{9x}{by}$.Since the curves intersect at right angles,the product of their slopes at the point of intersection $(x, y)$ is $-1$:$(\frac{3}{y}) 	imes (-\frac{9x}{by}) = -1$$\frac{27x}{by^2} = 1 \Rightarrow by^2 = 27x$.Substitute $y^2 = 6x$ into this equation:$b(6x) = 27x \Rightarrow 6b = 27 \Rightarrow b = \frac{27}{6} = \frac{9}{2}$.

If the curves $y^2=6x$ and $9x^2+by^2=16$ intersect each other at right angles,then the value of $b$ is

Similar Questions

The sum of the squares of the eccentricities of the conics $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$ is

How many parabolas can be drawn if the endpoints of the latus rectum are given?

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ are the same as the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2 = \dots$

Two mutually perpendicular tangents of the parabola $y^2 = 4ax$ meet the axis in $P_1$ and $P_2$. If $S$ is the focus of the parabola,then $\frac{1}{SP_1} + \frac{1}{SP_2}$ is equal to

The equation of the common tangent to the circle $x^{2}+y^{2}=2$ and the parabola $y^{2}=8x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module