Question:
Let a, \(\mathrm{b}\in\mathbf{R}\) and \(\mathrm{a}>0\). If the tangent at the point \((2,2)\) to the circle \(x^{2}+y^{2}=8\) touches the parabola \(y^{2}=4\mathrm{a}(x-\mathrm{b})\), then \(\mathrm{b}-\mathrm{a}\) is equal to ________.
Let a, \(\mathrm{b}\in\mathbf{R}\) and \(\mathrm{a}>0\). If the tangent at the point \((2,2)\) to the circle \(x^{2}+y^{2}=8\) touches the parabola \(y^{2}=4\mathrm{a}(x-\mathrm{b})\), then \(\mathrm{b}-\mathrm{a}\) is equal to ________.
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Find the tangent to the circle at (2, 2), then use the tangency condition \(c=a/m\) for the shifted parabola \(y^{2}=4a(x-b)\).
Updated On: Nov 12, 2026
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Correct Answer: 4
Solution and Explanation
Tangent to \(x^{2}+y^{2}=8\) at \((2,2)\): \(2x+2y=8\), i.e. \(x+y=4\).
Substitute \(y=4-x\) in \(y^{2}=4a(x-b)\): \((4-x)^{2}=4a(x-b)\Rightarrow x^{2}-(8+4a)x+16+4ab=0\).
Tangency means the discriminant is zero: \((8+4a)^{2}=4(16+4ab)\Rightarrow 64+64a+16a^{2}=64+16ab\Rightarrow 4a+a^{2}=ab\).
Since \(a>0\), dividing by \(a\): \(b=a+4\), so \(b-a=4\).
Answer: 4
Substitute \(y=4-x\) in \(y^{2}=4a(x-b)\): \((4-x)^{2}=4a(x-b)\Rightarrow x^{2}-(8+4a)x+16+4ab=0\).
Tangency means the discriminant is zero: \((8+4a)^{2}=4(16+4ab)\Rightarrow 64+64a+16a^{2}=64+16ab\Rightarrow 4a+a^{2}=ab\).
Since \(a>0\), dividing by \(a\): \(b=a+4\), so \(b-a=4\).
Answer: 4
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