JEE Advanced 2017 · Paper 1, Mathematics question 1

Find $a$. The line $y=2x+1$ touches $\dfrac{x^2}{a^2}-\dfrac{y^2}{16}=1$ when $c^2=a^2m^2-b^2$:
$$1=4a^2-16\implies a^2=\frac{17}4.$$

Test each triple (squares listed; a right triangle needs the largest square to equal the sum of the other two):

• (A) $a,4,1$: squares $\tfrac{17}4,16,1$. $\tfrac{17}4+1\ne16$. Not right-angled.
• (B) $a,4,2$: squares $\tfrac{17}4,16,4$. $\tfrac{17}4+4\ne16$. Not right-angled.
• (C) $2a,8,1$: squares $17,64,1$. $17+1\ne64$. Not right-angled.
• (D) $2a,4,1$: squares $17,16,1$. $16+1=17$. Right-angled.

So (A), (B) and (C) cannot be sides of a right triangle.

(A), (B), (C)