For the curve x^{2}+4xy+8y^{2}=64, the tangen | vedclass

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

Question

(B) Given curve is $x^{2}+4xy+8y^{2}=64$ ... $(i)$ Differentiating with respect to $x$, we get: $2x + 4(y + x\frac{dy}{dx}) + 16y\frac{dy}{dx} = 0$ $2

Vedclass · Accepted Answer

$(8, -4)$ and $(-8, 4)$

Vedclass · Answer

(B) Given curve is $x^{2}+4xy+8y^{2}=64$ ... $(i)$ Differentiating with respect to $x$, we get: $2x + 4(y + x\frac{dy}{dx}) + 16y\frac{dy}{dx} = 0$ $2x + 4y + (4x + 16y)\frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{x+2y}{2(x+4y)}$ Since the tangents are parallel to the $x$-axis, the slope $\frac{dy}{dx} = 0$. This implies $x + 2y = 0$, so $x = -2y$ ... (ii) Substituting $x = -2y$ into the original equation $(i)$: $(-2y)^{2} + 4(-2y)y + 8y^{2} = 64$ $4y^{2} - 8y^{2} + 8y^{2} = 64$ $4y^{2} = 64$ $y^{2} = 16 \Rightarrow y = \pm 4$ If $y = 4$, then $x = -2(4) = -8$. If $y = -4$, then $x = -2(-4) = 8$. Thus, the required points are $(-8, 4)$ and $(8, -4)$.

For the curve $x^{2}+4xy+8y^{2}=64$, the tangents are parallel to the $x$-axis only at the points

Similar Questions

Let $C$ be a curve $ax^2+2hxy+by^2+2gx+2fy+c=0$ in a Cartesian plane. By rotating the coordinate axes through an angle $\frac{\pi}{4}$ in the positive direction,if the transformed equation of $C$ is $Y^2+XY-X=0$,then $(h^2-ab)-2gf=$

The lines $y = -\frac{3}{2} x$ and $y = -\frac{2}{5} x$ intersect the curve $3x^2 + 4xy + 5y^2 - 4 = 0$ at the points $P$ and $Q$ respectively. The tangents drawn to the curve at $P$ and $Q$:

The equation $8 x^2+12 y^2-4 x+4 y-1=0$ represents

When does the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represent an ellipse?

If both roots of the equation $x^2 - 5ax + 6a = 0$ exceed $1$,then the range of '$a$' is

Vedclass Test Series

Exam Paper Generator

Online Exam Module