Tangents are drawn from the point P(3,4) to t | vedclass

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

Question

(B) $1.$ The equation of the chord of contact $AB$ for the point $P(3,4)$ with respect to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is given by $T=0

Vedclass · Accepted Answer

$(D, C, A)$

Vedclass · Answer

(B) $1.$ The equation of the chord of contact $AB$ for the point $P(3,4)$ with respect to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is given by $T=0$,i.e.,$\frac{3x}{9}+\frac{4y}{4}=1 \Rightarrow \frac{x}{3}+y=1 \Rightarrow x+3y-3=0$.To find the points of contact $A$ and $B$,we solve the system of equations: $x+3y=3$ and $\frac{x^2}{9}+\frac{y^2}{4}=1$. Substituting $x=3-3y$ into the ellipse equation: $\frac{(3-3y)^2}{9}+\frac{y^2}{4}=1 \Rightarrow (1-y)^2+\frac{y^2}{4}=1 \Rightarrow 1-2y+y^2+\frac{y^2}{4}=1 \Rightarrow \frac{5y^2}{4}-2y=0$. Thus,$y=0$ or $y=\frac{8}{5}$.If $y=0$,$x=3$. If $y=\frac{8}{5}$,$x=3-3(\frac{8}{5})=3-\frac{24}{5}=-\frac{9}{5}$. So,the points are $(3,0)$ and $(-\frac{9}{5}, \frac{8}{5})$. The correct option is $(D)$.$2.$ The orthocentre $H$ of $	riangle PAB$ is the intersection of altitudes. The line $AB$ is $x+3y-3=0$. The altitude from $P(3,4)$ to $AB$ has slope $3$ and passes through $(3,4)$,so $y-4=3(x-3) \Rightarrow y=3x-5$. The altitude from $A(3,0)$ to $PB$ (where $B=(-\frac{9}{5}, \frac{8}{5})$) has slope $m_{PB} = \frac{8/5-4}{-9/5-3} = \frac{-12/5}{-24/5} = \frac{1}{2}$. The altitude from $A$ is perpendicular to $PB$,so its slope is $-2$. Equation: $y-0=-2(x-3) \Rightarrow y=-2x+6$. Solving $3x-5=-2x+6 \Rightarrow 5x=11 \Rightarrow x=\frac{11}{5}$. Then $y=3(\frac{11}{5})-5 = \frac{33-25}{5} = \frac{8}{5}$. Thus,$H=(\frac{11}{5}, \frac{8}{5})$. The correct option is $(C)$.$3.$ The locus of a point equidistant from a fixed point $P$ and a fixed line $AB$ is a parabola with focus $P$ and directrix $AB$. The distance formula gives $(x-3)^2+(y-4)^2 = \frac{(x+3y-3)^2}{1^2+3^2}$. Expanding this: $10(x^2-6x+9+y^2-8y+16) = x^2+9y^2+9+6xy-6x-18y \Rightarrow 10x^2+10y^2-60x-80y+250 = x^2+9y^2+6xy-6x-18y+9 \Rightarrow 9x^2+y^2-6xy-54x-62y+241=0$. The correct option is $(A)$.

Similar Questions

If a focal chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ meets its minor axis at the point $(0,3)$,then the perpendicular distance from the centre of the ellipse to this focal chord is

The equation of the ellipse whose centre is $(2, -3)$,one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ is

For what values of $t$ does the line $x = at^2$ intersect the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at real points?

The equation of the ellipse whose distance between the foci is $8$ and the distance between the directrices is $18$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module