Question:
Line \(x+y+k=0\) touches hyperbola \(x^2-5y^2=5\). Point of contact is
Line \(x+y+k=0\) touches hyperbola \(x^2-5y^2=5\). Point of contact is
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For tangency, substitute line into curve and enforce discriminant = 0.
Updated On: Jun 22, 2026
- (5,2)
- (5,-2)
- \((-\frac{5}{2},\frac{1}{2})\)
- \((\frac{5}{2},-\frac{1}{2})\) \bigskip
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The Correct Option is D
Solution and Explanation
Concept:
For hyperbola tangent:
\[
xx_1/a^2 - yy_1/b^2 = 1
\]
Step 1: Standard form.
\[ \frac{x^2}{5}-\frac{y^2}{1}=1 \]
Step 2: Tangent condition.
\[ x+y+k=0 \Rightarrow y=-x-k \] Substitute in hyperbola and impose tangency condition.
Step 3: Point of contact.
Solving gives: \[ \left(\frac{5}{2},-\frac{1}{2}\right) \] \[ \boxed{(D)} \]
Step 1: Standard form.
\[ \frac{x^2}{5}-\frac{y^2}{1}=1 \]
Step 2: Tangent condition.
\[ x+y+k=0 \Rightarrow y=-x-k \] Substitute in hyperbola and impose tangency condition.
Step 3: Point of contact.
Solving gives: \[ \left(\frac{5}{2},-\frac{1}{2}\right) \] \[ \boxed{(D)} \]
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