Question:
A line parallel to the straight line \(2x - y = 0\) is tangent to the hyperbola
\(\frac{x^2}{4} - \frac{y^2}{2} = 1\)
at the point \((x_1, y_1)\). Then \(x_1^2 + 5y_1^2\) is equal to :
A line parallel to the straight line \(2x - y = 0\) is tangent to the hyperbola
\(\frac{x^2}{4} - \frac{y^2}{2} = 1\)
at the point \((x_1, y_1)\). Then \(x_1^2 + 5y_1^2\) is equal to :
\(\frac{x^2}{4} - \frac{y^2}{2} = 1\)
at the point \((x_1, y_1)\). Then \(x_1^2 + 5y_1^2\) is equal to :
Show Hint
The tangent has slope 2; use the tangent condition \(c^2 = a^2m^2 - b^2\) and the point of contact \((-a^2m/c,\,-b^2/c)\).
Updated On: Nov 5, 2026
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The Correct Option is B
Solution and Explanation
Here \(a^2 = 4,\ b^2 = 2\) and the tangent is parallel to \(y = 2x\), so \(m = 2\).
Tangent: \(y = 2x + c\) with \(c^2 = a^2m^2 - b^2 = 16 - 2 = 14\).
Point of contact: \(x_1 = -\frac{a^2 m}{c} = -\frac{8}{c}\), \(y_1 = -\frac{b^2}{c} = -\frac{2}{c}\).
\(x_1^2 = \frac{64}{14}\), \(y_1^2 = \frac{4}{14}\).
\(x_1^2 + 5y_1^2 = \frac{64 + 20}{14} = \frac{84}{14} = 6\).
Answer: (2) 6
Tangent: \(y = 2x + c\) with \(c^2 = a^2m^2 - b^2 = 16 - 2 = 14\).
Point of contact: \(x_1 = -\frac{a^2 m}{c} = -\frac{8}{c}\), \(y_1 = -\frac{b^2}{c} = -\frac{2}{c}\).
\(x_1^2 = \frac{64}{14}\), \(y_1^2 = \frac{4}{14}\).
\(x_1^2 + 5y_1^2 = \frac{64 + 20}{14} = \frac{84}{14} = 6\).
Answer: (2) 6
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