Question

If the normal to the rectangular hyperbola \[ x^2-y^2=1 \] at the point \(P\left(\frac{\pi}{4}\right)\) meets the curve again at \(Q(\theta)\), then \[ \sec^2\theta+\tan\theta= \]

• A. \(43\)
• B. \(57\)
• C. \(3\)
• D. \(1\)

Hint:

For the rectangular hyperbola \[ x^2-y^2=1, \] the parametric point is \[ (\sec\theta,\tan\theta) \] and the slope of normal can be found using implicit differentiation.

Answer 1

The Correct Option is B

Step 1: Use parametric form of rectangular hyperbola.
For \[ x^2-y^2=1, \] a point \(P(\theta)\) is represented as \[ (\sec\theta,\tan\theta) \] So, \[ P\left(\frac{\pi}{4}\right) = \left(\sec\frac{\pi}{4},\tan\frac{\pi}{4}\right) \] \[ = (\sqrt2,1) \]

Step 2: Find the normal at \(P(\sqrt2,1)\).
For \[ x^2-y^2=1, \] differentiate with respect to \(x\): \[ 2x-2y\frac{dy}{dx}=0 \] \[ \frac{dy}{dx}=\frac{x}{y} \] At \[ (\sqrt2,1), \] slope of tangent is \[ \sqrt2 \] Therefore, slope of normal is \[ -\frac{1}{\sqrt2} \] Equation of normal: \[ y-1=-\frac{1}{\sqrt2}(x-\sqrt2) \] \[ y=2-\frac{x}{\sqrt2} \]

Step 3: Find the second point of intersection.
From \[ y=2-\frac{x}{\sqrt2}, \] we get \[ x=\sqrt2(2-y) \] Substitute in \[ x^2-y^2=1 \] \[ \left[\sqrt2(2-y)\right]^2-y^2=1 \] \[ 2(2-y)^2-y^2=1 \] \[ 2(4-4y+y^2)-y^2=1 \] \[ 8-8y+2y^2-y^2=1 \] \[ y^2-8y+7=0 \] \[ (y-1)(y-7)=0 \] So, \[ y=1 \quad \text{or} \quad y=7 \] The point \(y=1\) is the original point \(P\), so for the second point \(Q\), \[ y=7 \] Then, \[ x=\sqrt2(2-7) \] \[ x=-5\sqrt2 \] Hence, \[ Q=(-5\sqrt2,7) \]

Step 4: Compare with parametric form of \(Q(\theta)\).
Since \[ Q(\theta)=(\sec\theta,\tan\theta), \] we get \[ \sec\theta=-5\sqrt2 \] and \[ \tan\theta=7 \] Therefore, \[ \sec^2\theta=50 \]

Step 5: Find the required value.
\[ \sec^2\theta+\tan\theta = 50+7 \] \[ =57 \]

Step 6: Final conclusion.
Hence, \[ \boxed{57} \] which corresponds to option (2).