Question

Given the equation of the ellipse $\frac{(x-3{)}^2}{16}+\frac{(y+4{)}^2}{49}=1$, a parabola is such that its vertex is the lowest point of the ellipse and it passes through the ends of the minor axis of the ellipse. The equation of the parabola is in the form $16y=a(x-h{)}^2-k$. Determine the value of $(a+h+k)$.

Answer 1

Answer: 186

Let $x-3=X$ and $y+4=Y$ hence the equation of the ellipse is $\frac{X^2}{16}+\frac{Y^2}{49}=1$ equation of a parabola with vertex on $A(0,-7)$ can be taken as
$X^2=\lambda (Y+7)$
It passes through $(4,0)$
$\therefore 16=7\lambda \Rightarrow \lambda =\frac{16}{7}$
Hence the equation of the parabola is new coordinate system is
$7X^2=16(Y+7)$
put $X=x-3$ and $Y=y+4$
$7(x-3{)}^2=16(y+11)$
$\therefore y=\frac{7}{16}(x-3{)}^2-11\text{ or }16y=7(x-3{)}^2-176$
$\text{ compare it with }16y=a(x-h{)}^2-k$
$a+h+k=7+3+176=186$