Question

Three normals are drawn from a point $P(1,1)$ to a parabola and let $A,B$ and $C$ be feet of perpendiculars with $A\equiv (0,0)$ and $B\equiv (3,-1)$. If equation of parabola is ${(x-a)}^2+{(y+b)}^2=\frac{{(x+cy-8)}^2}{10}$ then $a+2b+c$ is equal to

• A. $1$
• B. $3$
• C. $4$
• D. $7$

Answer 1

Answer: (A)

$1$

Parabola: $({x-a)}^2+({y+b)}^2=\frac{({x+cy-8)}^2}{10}............\left(1\right)$

Normals from $P(1,1)$ are drawn at $A,B$ and $C$

$A\equiv (0,0)$ and $B(3,-1)$ and let $C(h,k)$

We know that ordinates of foots of normal is when summed goes zero.

So, $C:(h,1)$

As $A$ and $B$ are on the parabola $=>a^2+b^2=\frac{16}{20}and9+a^2-6a+b^2+1-2b=\frac{c^2+25+10c}{10}$

$=>10-6a-2b+\frac{16}{10}=\frac{c^2+10c+25}{10}$

$=>100-60a-20b+16=c^2+10c+25...............\left(2\right)$

From $\left(1\right)$ $e=1$, $(a_1,b)$ is focus and $x=cy-8=0$ is directrix and $10=\sqrt{1+c^2}=>c^2+1=100$

Put $c^2=99$ in equation$\left(2\right)$

$=>10c+8+60a+20b=0$

$=>c+a+2b=\frac{8}{10}-5a$

We know that $\frac{2\left(SP\right)\left(SQ\right)}{SP+SQ}=semiLR$

$=>a=-\frac{1}{25}$

So, $c+a+2b=\frac{8}{10}-5(-\frac{1}{25})$

$=>a+2b+c=1.$