If the tangent at a point P on the parabola y | vedclass

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

Question

(D) For the parabola $y^2=3x$,the slope of the tangent at $P(x_1, y_1)$ is given by $2y \frac{dy}{dx} = 3$,so $\frac{dy}{dx} = \frac{3}{2y}$.Since the

Vedclass · Accepted Answer

$3\sqrt{5}$

Vedclass · Answer

(D) For the parabola $y^2=3x$,the slope of the tangent at $P(x_1, y_1)$ is given by $2y \frac{dy}{dx} = 3$,so $\frac{dy}{dx} = \frac{3}{2y}$.Since the tangent is parallel to $x+2y=1$ (slope $= -1/2$),we have $\frac{3}{2y_1} = -1/2$,which gives $y_1 = -3$. Substituting into $y^2=3x$,we get $x_1 = 3$. Thus,$P = (3, -3)$.For the ellipse $\frac{x^2}{4} + y^2 = 1$,the slope of the tangent at $(x, y)$ is $\frac{dy}{dx} = -\frac{x}{4y}$.The tangents at $Q$ and $R$ are perpendicular to $x-y=2$ (slope $= 1$),so their slope is $-1$. Thus,$-\frac{x}{4y} = -1$,which implies $x = 4y$.Substituting $x=4y$ into the ellipse equation: $\frac{(4y)^2}{4} + y^2 = 1 \Rightarrow 4y^2 + y^2 = 1 \Rightarrow 5y^2 = 1 \Rightarrow y = \pm \frac{1}{\sqrt{5}}$.Then $x = \pm \frac{4}{\sqrt{5}}$. So $Q = (\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $R = (-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}})$.The area of $	riangle PQR = \frac{1}{2} |x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)|$.Area $= \frac{1}{2} |3(\frac{1}{\sqrt{5}} - (-\frac{1}{\sqrt{5}})) + \frac{4}{\sqrt{5}}(-\frac{1}{\sqrt{5}} - (-3)) + (-\frac{4}{\sqrt{5}})(-3 - \frac{1}{\sqrt{5}})|$.Area $= \frac{1}{2} |3(\frac{2}{\sqrt{5}}) + \frac{4}{\sqrt{5}}(3 - \frac{1}{\sqrt{5}}) - \frac{4}{\sqrt{5}}(-3 - \frac{1}{\sqrt{5}})| = \frac{1}{2} |\frac{6}{\sqrt{5}} + \frac{12}{\sqrt{5}} - \frac{4}{5} + \frac{12}{\sqrt{5}} + \frac{4}{5}| = \frac{1}{2} |\frac{30}{\sqrt{5}}| = 3\sqrt{5}$.

If the tangent at a point $P$ on the parabola $y^2=3x$ is parallel to the line $x+2y=1$ and the tangents at the points $Q$ and $R$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ are perpendicular to the line $x-y=2$,then the area of the triangle $PQR$ is:

Similar Questions

The equations of the pairs of opposite sides of a parallelogram are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$. The equations of its diagonals are:

If the tangent to $y^{2}=4ax$ at the point $(at^{2}, 2at)$ where $|t|>1$ is a normal to $x^{2}-y^{2}=a^{2}$ at the point $(a \sec \theta, a \tan \theta)$, then

If the normal to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ at the point $\left( 2, \frac{3}{2} \right)$ is tangent to a parabola,find the equation of the parabola.

Find the angle of intersection of the curves $y^{2}=4ax$ and $x^{2}=4by$.

Let the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module