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:
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:
Show 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.
Updated On: Jun 10, 2026
- \(\pm \dfrac{3\sqrt2}{4}\)
- \(\pm \dfrac{4\sqrt2}{3}\)
- \(\pm \dfrac{5\sqrt3}{4}\)
- \(\pm \dfrac{3\sqrt3}{2}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For the hyperbola
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,
\]
the conjugate hyperbola is
\[
\frac{y^2}{b^2}-\frac{x^2}{a^2}=1.
\]
The ends of the transverse axis of the conjugate hyperbola are
\[
(0,b)
\quad\text{and}\quad
(0,-b).
\]
A circle having these two points as endpoints of a diameter can be obtained directly. After forming the circle, we find a tangent which is common to both the circle and the hyperbola. Equating the tangent conditions gives the required slope.
Step 1: Identify \(a\) and \(b\). Given hyperbola: \[ \frac{x^2}{25}-\frac{y^2}{16}=1. \] Comparing with \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] we obtain \[ a=5, \qquad b=4. \]
Step 2: Write the conjugate hyperbola. The conjugate hyperbola is \[ \frac{y^2}{16}-\frac{x^2}{25}=1. \] Its transverse-axis endpoints are \[ B(0,4), \qquad B'(0,-4). \]
Step 3: Form the circle having \(BB'\) as diameter. The midpoint of \(BB'\) is \[ (0,0). \] The radius is \[ 4. \] Therefore the circle is \[ x^2+y^2=16. \]
Step 4: Take a common tangent with slope \(m\). Let the common tangent be \[ y=mx+c. \] For the circle \[ x^2+y^2=16, \] the condition of tangency is \[ c^2=16(1+m^2). \] \[ c=\pm4\sqrt{1+m^2}. \]
Step 5: Use tangency condition for the hyperbola. For \[ \frac{x^2}{25}-\frac{y^2}{16}=1, \] a tangent of slope \(m\) is \[ y=mx\pm\sqrt{25m^2-16}. \] Thus \[ c^2=25m^2-16. \]
Step 6: Equate the two values of \(c^2\). Since the same line is tangent to both curves, \[ 16(1+m^2)=25m^2-16. \] Expanding, \[ 16+16m^2=25m^2-16. \] \[ 32=9m^2. \] \[ m^2=\frac{32}{9}. \] \[ m=\pm\frac{4\sqrt2}{3}. \]
Step 7: Final Conclusion. Therefore the slope of the common tangent is \[ \boxed{\pm\frac{4\sqrt2}{3}}. \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]
Step 1: Identify \(a\) and \(b\). Given hyperbola: \[ \frac{x^2}{25}-\frac{y^2}{16}=1. \] Comparing with \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] we obtain \[ a=5, \qquad b=4. \]
Step 2: Write the conjugate hyperbola. The conjugate hyperbola is \[ \frac{y^2}{16}-\frac{x^2}{25}=1. \] Its transverse-axis endpoints are \[ B(0,4), \qquad B'(0,-4). \]
Step 3: Form the circle having \(BB'\) as diameter. The midpoint of \(BB'\) is \[ (0,0). \] The radius is \[ 4. \] Therefore the circle is \[ x^2+y^2=16. \]
Step 4: Take a common tangent with slope \(m\). Let the common tangent be \[ y=mx+c. \] For the circle \[ x^2+y^2=16, \] the condition of tangency is \[ c^2=16(1+m^2). \] \[ c=\pm4\sqrt{1+m^2}. \]
Step 5: Use tangency condition for the hyperbola. For \[ \frac{x^2}{25}-\frac{y^2}{16}=1, \] a tangent of slope \(m\) is \[ y=mx\pm\sqrt{25m^2-16}. \] Thus \[ c^2=25m^2-16. \]
Step 6: Equate the two values of \(c^2\). Since the same line is tangent to both curves, \[ 16(1+m^2)=25m^2-16. \] Expanding, \[ 16+16m^2=25m^2-16. \] \[ 32=9m^2. \] \[ m^2=\frac{32}{9}. \] \[ m=\pm\frac{4\sqrt2}{3}. \]
Step 7: Final Conclusion. Therefore the slope of the common tangent is \[ \boxed{\pm\frac{4\sqrt2}{3}}. \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]
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