If two curves x^2-4y^2=2 and 8x^2=40-my^2 are | vedclass

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

Question

(B) Given curves are $C_1: x^2-4y^2=2$ and $C_2: 8x^2+my^2=40$. Differentiating $C_1$ with respect to $x$: $2x - 8y \frac{dy}{dx} = 0 \implies \frac{d

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) Given curves are $C_1: x^2-4y^2=2$ and $C_2: 8x^2+my^2=40$. Differentiating $C_1$ with respect to $x$: $2x - 8y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{4y} = m_1$. Differentiating $C_2$ with respect to $x$: $16x + 2my \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{8x}{my} = m_2$. Since the curves are orthogonal,$m_1 	imes m_2 = -1$. $(\frac{x}{4y}) 	imes (-\frac{8x}{my}) = -1 \implies \frac{8x^2}{4my^2} = 1 \implies 2x^2 = my^2$. From $C_1$,$x^2 = 2 + 4y^2$. Substituting this into the condition: $2(2 + 4y^2) = my^2 \implies 4 + 8y^2 = my^2 \implies (m-8)y^2 = 4$. For the curves to intersect at a point $(x, y)$,we solve the system: $x^2 - 4y^2 = 2$ and $8x^2 + my^2 = 40$. Multiply the first by $8$: $8x^2 - 32y^2 = 16$. Subtract from the second: $(m+32)y^2 = 24 \implies y^2 = \frac{24}{m+32}$. Substitute $y^2$ into $x^2 = 2 + 4y^2$: $x^2 = 2 + \frac{96}{m+32} = \frac{2m+64+96}{m+32} = \frac{2m+160}{m+32}$. Substitute into $2x^2 = my^2$: $2(\frac{2m+160}{m+32}) = m(\frac{24}{m+32}) \implies 4m + 320 = 24m \implies 20m = 320 \implies m = 16$.

If two curves $x^2-4y^2=2$ and $8x^2=40-my^2$ are orthogonal to each other,then $m=$

Similar Questions

Let $P$ be a moving point on the circle $x^2 + y^2 - 6x - 8y + 21 = 0$. Then, the maximum distance of $P$ from the vertex of the parabola $x^2 + 6x + y + 13 = 0$ is equal to:

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 $b^2 = \dots$

The locus of the midpoints of the chords of the hyperbola $x^{2}-y^{2}=4$,which touch the parabola $y^{2}=8x$,is:

If the curves $\frac{x^2}{4}+\frac{y^2}{9}=1$ and $\frac{x^2}{16}-\frac{y^2}{k}=1$ cut each other orthogonally, then $k=$

If for $\theta \in \left[-\frac{\pi}{3}, 0\right]$,the points $(x, y) = \left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $xy+\alpha x+\beta y+\gamma=0$,then $\alpha^2+\beta^2+\gamma^2$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module