A conic passes through the point (2, 4) and i | vedclass

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

Question

(C) Let the point of tangency be $(x, y)$ and the slope of the tangent be $m = \frac{dy}{dx}$.The equation of the tangent is $Y - y = m(X - x)$.The $X

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the point of tangency be $(x, y)$ and the slope of the tangent be $m = \frac{dy}{dx}$.The equation of the tangent is $Y - y = m(X - x)$.The $X$-intercept (where $Y=0$) is $X = x - \frac{y}{m}$.The $Y$-intercept (where $X=0$) is $Y = y - mx$.Since the point $(x, y)$ bisects the segment between the axes,we have $x = \frac{1}{2}(x - \frac{y}{m} + 0)$ $\Rightarrow 2x = x - \frac{y}{m}$ $\Rightarrow \frac{y}{m} = -x$.Substituting $m = \frac{dy}{dx}$,we get $y \frac{dx}{dy} = -x \Rightarrow \frac{dy}{y} = -\frac{dx}{x}$.Integrating both sides,$\ln y = -\ln x + C \Rightarrow xy = c$.Since the conic passes through $(2, 4)$,$c = 2 	imes 4 = 8$. Thus,the equation is $xy = 8$.This is a rectangular hyperbola $xy = c^2/2$ with $c^2/2 = 8 \Rightarrow c^2 = 16$.The standard form is $X'Y' = a^2/2$ where $X', Y'$ are axes rotated by $45^\circ$.The foci of $xy = c^2/2$ are $(\pm c, \pm c)$.Here $c^2/2 = 8$ $\Rightarrow c^2 = 16$ $\Rightarrow c = 4$.Thus,the foci are $(4, 4)$ and $(-4, -4)$.

$A$ conic passes through the point $(2, 4)$ and is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangency. Then the foci of the conic are:

Similar Questions

The angle between the curves $x^2-y^2=4$ and $x^2+y^2=4\sqrt{2}$ is

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

If the tangent drawn to the parabola $y^2=4x$ at $(t^2, 2t)$ is the normal to the ellipse $4x^2+5y^2=20$ at $(\sqrt{5} \cos \theta, 2 \sin \theta)$,then

If $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$ and the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ respectively,and $(e_{1}, e_{2})$ is a point on the ellipse $15x^{2}+3y^{2}=k$,then $k$ is equal to:

If the ellipse $4x^2 + 9y^2 = 36$ is confocal with a hyperbola whose length of the transverse axis is $2$,then the points of intersection of the ellipse and hyperbola lie on the circle:

Vedclass Test Series

Exam Paper Generator

Online Exam Module