Question

Consider the hyperbola \[ S \equiv \frac{x^2}{25}-\frac{y^2}{16}-1=0. \] Let \(B,B'\) be the ends of the transverse axis of the conjugate hyperbola of \(S=0\). If \(C\) is the circle with \(B,B'\) as ends of a diameter, then the slope of a common tangent to \(C\) and the given hyperbola is:

• A. \(\pm \dfrac{3\sqrt2}{4}\)
• B. \(\pm \dfrac{4\sqrt2}{3}\)
• C. \(\pm \dfrac{5\sqrt3}{4}\)
• D. \(\pm \dfrac{3\sqrt3}{2}\)

Hint:

For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the tangent with slope \(m\) is \[ y=mx\pm\sqrt{a^2m^2-b^2}. \] This formula is extremely useful in slope-based tangent problems.

Answer 1

The Correct Option is B

Step 1: Read the hyperbola.
The hyperbola is $\dfrac{x^2}{25}-\dfrac{y^2}{16}=1$, so $a^2=25$ and $b^2=16$, giving $a=5,\,b=4$.

Step 2: Find the conjugate hyperbola's transverse-axis ends.
The conjugate hyperbola is $\dfrac{y^2}{16}-\dfrac{x^2}{25}=1$. Its transverse axis is along the $y$-axis, with ends $B(0,4)$ and $B'(0,-4)$.

Step 3: Build the circle.
The circle $C$ has $B$ and $B'$ as a diameter, so its centre is the origin and radius $4$. Thus $C:\,x^2+y^2=16$.

Step 4: Write the tangent conditions.
A line $y=mx+c$ touches the circle $x^2+y^2=16$ when $c^2=16(1+m^2)$. It touches the hyperbola $\dfrac{x^2}{25}-\dfrac{y^2}{16}=1$ when $c^2=25m^2-16$.

Step 5: Equate the two conditions.
\[ 16(1+m^2)=25m^2-16. \] \[ 16+16m^2=25m^2-16. \] \[ 32=9m^2 \quad\Rightarrow\quad m^2=\frac{32}{9}. \]

Step 6: Solve for the slope.
\[ m=\pm\sqrt{\frac{32}{9}}=\pm\frac{4\sqrt2}{3}. \] This is option 2.
\[ \boxed{\pm\dfrac{4\sqrt2}{3}} \]