Question:
A differential equation representing the family of parabolas with axis parallel to $\mathrm{y}$-axis and whose length of latus rectum is the distance of the point $(2,-3)$ form the line $3 x+4 y=5$, is given by :
Correct Option: , 4
Solution:
$\alpha . R=\frac{|3(2)+4(-3)-5|}{5}=\frac{11}{5}$
$(x-h)^{2}=\frac{11}{5}(y-k)$
differentiate w.r.t ' $x$ ' : $-$.
$2(x-h)=\frac{11}{5} \frac{d y}{d x}$
again differentiate
$2=\frac{11}{5} \frac{d^{2} y}{d x^{2}}$
$\frac{11 \mathrm{~d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=10$
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