Let the image of parabola x^2=4y, in the line x-y=1 be (y+a)^2=b(x-c), a,b,c N.

Question

Let the image of parabola ${x}^{2}=4y$, in the line $x-y=1$ be $(y+a{)}^{2}=b(x-c)$, $a,b,c\in N$. Then $a+b+c$ is equal to

Accepted Answer

Let parametric point on parabola ${x}^{2}=4y$ is $(2t,{t}^{2})$.

Key concept: image of point $(h,k)$ in the line $ax+by+c=0$ is given as $\frac{x-h}{a}=\frac{y-k}{b}=\frac{-2(ah+bk+c)}{{a}^{2}+{b}^{2}}$

So, Image of point in line $x-y-1=0$ is given as

$\frac{x-2t}{1}=\frac{y-{t}^{2}}{-1}=\frac{-2(2t-{t}^{2}-1)}{{1}^{2}+(-1{)}^{2}}$

equating

$\begin{gathered} \frac{x-2t}{1}=\frac{-2(2t-{t}^{2}-1)}{2} \ x=-2t+{t}^{2}+1+2t\Rightarrow x={t}^{2}+1 \end{gathered}$

equating

$\begin{gathered} \frac{y-{t}^{2}}{-1}=\frac{-2(2t-{t}^{2}-1)}{2} \ y-{t}^{2}=2t-{t}^{2}-1\Rightarrow y=2t-1 \end{gathered}$

from (1) and (2) eliminate parameter t to get

image of curve.

so, from (2) put $t=\frac{y+1}{2}$ in equation (1).

$\begin{gathered} x={(\frac{y+1}{2})}^{2}+1 \ \frac{(y+1{)}^{2}}{4}=x-1 \ (y+1{)}^{2}=4(x-1) \end{gathered}$

Compare with given image of curve $(y+a{)}^{2}=b(x-c)$

to find value of $a,b,\&c$.

$\begin{gathered} a=1,b=4,c=1 \ a+b+c=1+4+1=6. \end{gathered}$

Solutions · 1