1.
If $$a>2b>0$$ then the positive value of $$m$$ for which $$y = mx - b\sqrt {1 + {m^2}} $$ is a common tangent to $${x^2} + {y^2} = {b^2}$$ and $${\left( {x - a} \right)^2} + {y^2} = {b^2}$$ is :
$${x^2} - 5xy + 6{y^2} = 0$$ represents a pair of straight lines given by $$x-3y=0$$ and $$x-2y=0.$$
Also $$a{x^2} + b{y^2} + c = 0$$ will represent a circle if $$a=b$$ and $$c$$ is of sign opposite to that of $$a.$$
5.
The line joining $$\left( {5,\,0} \right)$$ to $$\left( {10\,\cos \,\theta ,\,10\,\sin \,\theta } \right)$$ is divided internally in the ratio $$2 : 3$$ at $$P$$. If $$\theta $$ varies, then the locus of $$P$$ is :
Let $$P\left( {x,\,y} \right)$$ be the point dividing the join of $$A$$ and $$B$$ in the ratio $$2 : 3$$ internally, then
$$\eqalign{
& x = \frac{{20\,\cos \,\theta + 15}}{5} = 4\,\cos \,\theta + 3 \cr
& \Rightarrow \cos \,\theta = \frac{{x - 3}}{4}......\left( {\text{i}} \right) \cr
& y = \frac{{20\,\sin \,\theta + 0}}{5} = 4\,\sin \,\theta \cr
& \Rightarrow \sin \,\theta = \frac{y}{4}......\left( {{\text{ii}}} \right) \cr} $$
Squaring and adding $$\left( {{\text{i}}} \right)$$ and $$\left( {{\text{ii}}} \right),$$ we get the required locus $${\left( {x - 3} \right)^2} + {y^2} = 16,$$ which is a circle.
6.
The locus of the vertices of the family of parabolas $$y = \frac{{{a^3}{x^2}}}{3} + \frac{{{a^2}x}}{2} - 2a$$ is :
Given equation of ellipse can be written as
$$\frac{{{x^2}}}{6} + \frac{{{y^2}}}{2} = 1\,\,\,\,\,\,\, \Rightarrow {a^2} = 6,\,\,\,{b^2} = 2$$
Now, equation of any variable tangent is
$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} .....({\text{i}})$$
where $$m$$ is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
$$y = \frac{{ - x}}{m}.....({\text{ii}})$$
Eliminating $$m,$$ we get
$$\eqalign{
& \left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2} \cr
& \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2} \cr
& \Rightarrow \boxed{{{\left( {{x^2} + {y^2}} \right)}^2} = 6{x^2} + 2{y^2}} \cr} $$
8.
The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0,$$ at $$\left( {1,\,1} \right)$$
A.
meets the curve again in the third quadrant.
B.
meets the curve again in the fourth quadrant.
C.
does not meet the curve again.
D.
meets the curve again in the second quadrant.
Answer :
meets the curve again in the fourth quadrant.
Given curve is
$${x^2} + 2xy - 3{y^2} = 0.....(1)$$
Differentiate w.r.t. $$x,\,\,\,2x + 2x\frac{{dy}}{{dx}} + 2y - 6y\frac{{dy}}{{dx}} = 0$$
$${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,\,1} \right)}} = 1$$
Equation of normal at $$\left( {1,\,1} \right)$$ is
$$y = 2 - x.....(2)$$
Solving equation (1) and (2), we get $$x=1 ,\,3$$
Point of intersection $$\left( {1,\,1} \right),\,\left( {3,\, - 1} \right)$$
Normal cuts the curve again in 4th quadrant.
9.
The locus of the orthocenter of the triangle formed by the
lines
$$\eqalign{
& \left( {1 + p} \right)x - py + p\left( {1 + p} \right) = 0, \cr
& \left( {1 + q} \right)x - qy + q\left( {1 + q} \right) = 0, \cr} $$
and $$y=0,$$ where $$p \ne q,$$ is :