The set of all values of a^2 for which the li | vedclass

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

Question

$x ^2+ y ^2-\frac{(1+ a ) x }{2}-\frac{(1- a ) y }{2}=0$

$	ext { Centre }\left(\frac{1+ a }{4}, \frac{1- a }{4}ight) \Rightarrow( h , k )$

$P

Vedclass · Accepted Answer

$(8, \infty)$

Vedclass · Answer

$x ^2+ y ^2-\frac{(1+ a ) x }{2}-\frac{(1- a ) y }{2}=0$ $ ext { Centre }\left(\frac{1+ a }{4}, \frac{1- a }{4} ight) \Rightarrow( h , k )$ $P \left(\frac{1+ a }{2}, \frac{1- a }{2} ight) \Rightarrow(2 h , 2 k )$ Equation of chord $\Rightarrow T=S_1$ $\Rightarrow( x - y ) \lambda-\frac{2 h ( x +\lambda)}{2}-\frac{(2 k )( y -\lambda)}{2}$ $=2 \lambda^2-2 h (\lambda)+2 k \lambda$ Now, $\lambda(2 h , 2 k )$ satisfies the chord $ herefore(2 h -2 k ) \lambda- h ( x +\lambda)- k ( y -\lambda)$ $\Rightarrow 2 \lambda^2+4 k \lambda-4 h \lambda+ h \lambda- k \lambda+ hx + ky =0$ $\Rightarrow 2 \lambda^2+\lambda(3 k -3 h )+ ky + hx =0$ $\Rightarrow D > 0$ $\Rightarrow 9( k - h )^2-8( ky + hx ) > 0$ $\Rightarrow 9( k - h )^2-8\left(2 k ^2+2 h ^2 ight) > 0$ $\Rightarrow-7 k ^2-7 h ^2-18 kh > 0$ $\Rightarrow 7 k ^2+7 h ^2+18 kh < 0$ $\Rightarrow 7\left(\frac{1- a }{4} ight)^2+7\left(\frac{1+ a }{4} ight)^2+18\left(\frac{1- a ^2}{16} ight) < 0$ $\Rightarrow 7\left[\frac{2\left(1+ a ^2 ight)}{16} ight]+\frac{18\left(1- a ^2 ight)}{16} < 0, \quad a ^2= t$ $\Rightarrow \frac{7}{8}(1+ t )+\frac{18(1- t )}{16} < 0$ $\Rightarrow \frac{14+14 t +18-18 t }{16} < 0$ $\Rightarrow 4 t > 32$ $t > 8 \quad a ^2 > 8$

The set of all values of $a^2$ for which the line $x + y =0$ bisects two distinct chords drawn from a point $P\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ on the circle $2 x ^2+2 y ^2-(1+ a ) x -(1- a ) y =0$ is equal to:

Similar Questions

Tangents $AB$ and $AC$ are drawn from the point $A(0,\,1)$ to the circle ${x^2} + {y^2} - 2x + 4y + 1 = 0$. Equation of the circle through $A, B$ and $C$ is

Equation of the tangent to the circle, at the point $(1 , -1)$ whose centre is the point of intersection of the straight lines $x - y = 1$ and $2x + y= 3$ is

A tangent drawn from the point $(4, 0)$ to the circle $x^2 + y^2 = 8$ touches it at a point $A$ in the first quadrant. The co-ordinates of another point $B$ on the circle such that $l\, (AB) = 4$ are :

If a line, $y=m x+c$ is a tangent to the circle, $(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $\mathrm{L}_{1},$ where $\mathrm{L}_{1}$ is the tangent to the circle, $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then

The equation of circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$, where $c$ is