JEE Mains · Maths · STD 12 - 9. differential equations
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 :
- A \(10 \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=11\)
- B \(11 \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=10\)
- C \(10 \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=11\)
- D \(11 \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=10\)
Answer & Solution
Correct Answer
(B) \(11 \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=10\)
Step-by-step Solution
Detailed explanation
\(\alpha . \mathrm{R}=\frac{|3(2)+4(-3)-5|}{5}=\frac{11}{5}\) \((x-h)^{2}=\frac{11}{5}(y-k)\) differentiate w.r.t ' \(x^{\prime}\) : - \(2(x-h)=\frac{11}{5} \frac{d y}{d x}\) again differentiate \(2=\frac{11}{5} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\)…
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