Question

The length of the parabola \(y^2=12x\) cut off by the latus-rectum is

• A. \(6\left(\sqrt{2}+\log(1+\sqrt{2})\right)\)
• B. \(3\left(\sqrt{2}+\log(1+\sqrt{2})\right)\)
• C. \(6\left(\sqrt{2}-\log(1+\sqrt{2})\right)\)
• D. \(3\left(\sqrt{2}-\log(1+\sqrt{2})\right)\)

Hint:

For \(y^2=4ax\), latus rectum endpoints are \((a,\pm 2a)\). Use symmetry in arc length integrals to simplify calculations.

Answer 1

The Correct Option is A

Step 1: Compare with standard parabola form.
Standard form:
\[ y^2=4ax \]
Given:
\[ y^2=12x \Rightarrow 4a=12 \Rightarrow a=3 \]
Step 2: End points of latus rectum.
For \(y^2=4ax\), latus rectum is line \(x=a\) and endpoints are:
\[ (a,2a),\ (a,-2a) \]
So endpoints are:
\[ (3,6),\ (3,-6) \]
Step 3: Express curve in terms of \(y\).
\[ x=\frac{y^2}{12} \Rightarrow \frac{dx}{dy}=\frac{2y}{12}=\frac{y}{6} \]
Step 4: Arc length formula with respect to \(y\).
\[ L=\int_{-6}^{6}\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy =\int_{-6}^{6}\sqrt{1+\left(\frac{y}{6}\right)^2}\,dy \]
Step 5: Simplify integral using symmetry.
\[ L=2\int_{0}^{6}\sqrt{1+\frac{y^2}{36}}\,dy =2\int_{0}^{6}\sqrt{\frac{36+y^2}{36}}\,dy \]
\[ = \frac{2}{6}\int_{0}^{6}\sqrt{36+y^2}\,dy = \frac{1}{3}\int_{0}^{6}\sqrt{36+y^2}\,dy \]
Step 6: Use standard integral.
\[ \int\sqrt{y^2+a^2}\,dy=\frac{y}{2}\sqrt{y^2+a^2}+\frac{a^2}{2}\ln\left|y+\sqrt{y^2+a^2}\right| \]
Here \(a^2=36\Rightarrow a=6\).
So:
\[ \int_{0}^{6}\sqrt{36+y^2}\,dy = \left[\frac{y}{2}\sqrt{y^2+36}+18\ln\left(y+\sqrt{y^2+36}\right)\right]_{0}^{6} \]
At \(y=6\):
\[ \frac{6}{2}\sqrt{72} + 18\ln(6+\sqrt{72}) =3\cdot 6\sqrt{2}+18\ln(6+6\sqrt2) =18\sqrt2 +18\ln(6(1+\sqrt2)) \]
At \(y=0\):
\[ 0 + 18\ln(6) \]
Subtracting:
\[ =18\sqrt2 +18\ln(6(1+\sqrt2)) -18\ln 6 \]
\[ =18\sqrt2 +18\ln(1+\sqrt2) \]
Step 7: Multiply by \(\frac{1}{3}\).
\[ L=\frac{1}{3}\left(18\sqrt2 +18\ln(1+\sqrt2)\right) =6\left(\sqrt2+\ln(1+\sqrt2)\right) \]
Final Answer:
\[ \boxed{6\left(\sqrt{2}+\log(1+\sqrt{2})\right)} \]