Consider the conic C: 25(x - 1)^2 + 25(y + 1) | vedclass

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

Question

(B) The given equation of the conic $C$ is $25((x - 1)^2 + (y + 1)^2) = (3x - 4y)^2$. Dividing by $25$,we get $(x - 1)^2 + (y + 1)^2 = \left(\frac{3x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The given equation of the conic $C$ is $25((x - 1)^2 + (y + 1)^2) = (3x - 4y)^2$. Dividing by $25$,we get $(x - 1)^2 + (y + 1)^2 = \left(\frac{3x - 4y}{5}\right)^2$. This is in the form $SP^2 = PM^2$,where $S = (1, -1)$ is the focus and $3x - 4y = 0$ is the directrix. Since the eccentricity $e = 1$,the conic $C$ is a parabola. The locus of the point of intersection of perpendicular tangents to a parabola is its directrix. Thus,the curve $E$ is the line $3x - 4y = 0$. The minimum distance between the point $(2, -1)$ and the line $3x - 4y = 0$ is the perpendicular distance $d = \frac{|3(2) - 4(-1)|}{\sqrt{3^2 + (-4)^2}}$. $d = \frac{|6 + 4|}{\sqrt{9 + 16}} = \frac{10}{5} = 2$.

Consider the conic $C: 25(x - 1)^2 + 25(y + 1)^2 = (3x - 4y)^2$. If the curve $E$ is the locus of the point of intersection of perpendicular tangents to the conic $C$,then the minimum distance between the curve $E$ and the point $(2, -1)$ is:

Similar Questions

Let a focal chord $12x + 5y - 27 = 0$ of the parabola $y^2 = kx$ intersect the parabola at the points $P$ and $P^{\prime}$. If $S$ is the focus of this parabola,then $9(SP + SP^{\prime}) = $

The line $y=mx+3$ is tangent to the parabola $y^2=4x$,if the value of $m$ is

If the vertices of a triangle are $(am_1^2, 2am_1), (am_2^2, 2am_2),$ and $(am_3^2, 2am_3),$ then the area of the triangle is

If $x = t^2$ and $y = 2t$,then the equation of the normal at $t = 1$ is

Let the normal at the point $P$ on the parabola $y^{2} = 6x$ pass through the point $(5, -8)$. If the tangent at $P$ to the parabola intersects its directrix at the point $Q$,then the ordinate of the point $Q$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module