Hyperbola
Tangent from External Point
Grade 11
Question:
<p>The tangent to the hyperbola <i>y</i> = <i>(x + 9)/(x − 5)</i> passing through the origin is</p>
<p>(a) <i>x</i> − 25<i>y</i> = 0</p>
<p>(b) 5<i>x</i> − <i>y</i> = 0</p>
<p>(c) 5<i>x</i> + <i>y</i> = 0</p>
<p>(d) <i>x</i> + 25<i>y</i> = 0</p>
Step-by-Step Solution
Key Concept: A tangent line from the origin to the curve must satisfy both the tangent condition (at point of tangency) and the pass-through condition (origin lies on the tangent).
<p>Here, <i>y</i> = 1 + 4/(<i>x</i> − 5) ⟹ d<i>y</i>/d<i>x</i> at (<i>x</i><sub>1</sub>, <i>y</i><sub>1</sub>) = −4/(<i>x</i><sub>1</sub> − 5)<sup>2</sup></p><p>Now, equation of tangent is</p><p><i>y</i> − (1 + 4/(<i>x</i><sub>1</sub> − 5)) = −4/(<i>x</i><sub>1</sub> − 5)<sup>2</sup> · (<i>x</i> − <i>x</i><sub>1</sub>)</p><p>Since it passes through (0, 0), therefore</p><p>−(1 + 4/(<i>x</i><sub>1</sub> − 5)) = −4<i>x</i><sub>1</sub>/(<i>x</i><sub>1</sub> − 5)<sup>2</sup></p><p>⟹ −(<i>x</i><sub>1</sub> − 5)<sup>2</sup> − 4(<i>x</i><sub>1</sub> − 5) = 4<i>x</i><sub>1</sub></p><p>⟹ <i>x</i><sub>1</sub><sup>2</sup> − 18<i>x</i><sub>1</sub> + 45 = 0</p><p>⟹ (<i>x</i><sub>1</sub> − 15)(<i>x</i><sub>1</sub> − 3) = 0 ⟹ <i>x</i><sub>1</sub> = −15 or −3</p><p>So, equation of tangents are <i>x</i> − 25<i>y</i> = 0 or <i>x</i> − <i>y</i> = 0</p>
Correct Answer: A