NIMCET 2022 — Mathematics PYQ

1. The equations of the two curves are given as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2-y^2=c^2$. 
2. The derivative of the first curve with respect to $x$ is found by implicit differentiation: $\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$, which gives $\frac{dy}{dx}=-\frac{b^{2}x}{a^{2}y}$. This is denoted as $m_1$. 
3. The derivative of the second curve with respect to $x$ is found by implicit differentiation: $2x-2y\frac{dy}{dx}=0$, which gives $\frac{dy}{dx}=\frac{x}{y}$. This is denoted as $m_2$. 
4. Since the curves intersect at right angles, the product of their slopes at the point of intersection $(x,y)$ must be $-1$, i.e., $m_1{m}_2=-1$. 
5. Substituting the expressions for $m_1$ and $m_2$, we get $\left(-\frac{b^{2}x}{a^{2}y}\right)\left(\frac{x}{y}\right)=-1$. 
6. This simplifies to $-\frac{b^2{x}^2}{a^2{y}^2}=-1$, or $\frac{b^2{x}^2}{a^2{y}^2}=1$. 
7. Rearranging this equation gives $b^2{x}^2=a^2{y}^2$. 
8. From the second curve's equation, $x^2-y^2=c^2$, we can express $x^2=c^2+y^2$. 
9. Substitute this expression for $x^2$ into $b^2{x}^2=a^2{y}^2$: $b^2(c^2+y^2)=a^2{y}^2$. 
10. Expanding and rearranging gives $b^2{c}^2+b^2{y}^2=a^2{y}^2$, which leads to $b^2{c}^2=(a^2-b^2)y^2$. [5]  
11. From the first curve's equation, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we can express $y^2=b^2\left(1-\frac{x^2}{a^2}\right)$. [3]  
12. Substitute this expression for $y^2$ into $b^2{x}^2=a^2{y}^2$: $b^2{x}^2=a^2\left(b^2\left(1-\frac{x^2}{a^2}\right)\right)$. [3]  
13. This simplifies to $b^2{x}^2=a^2{b}^2-b^2{x}^2$. 
14. Rearranging gives $2b^2{x}^2=a^2{b}^2$, which implies $x^2=\frac{a^2}{2}$. 
15. Substitute this value of $x^2$ into $x^2-y^2=c^2$: $\frac{a^2}{2}-y^2=c^2$. 
16. This gives $y^2=\frac{a^2}{2}-c^2$. [3]  
17. Now, substitute the expressions for $x^2$ and $y^2$ into $b^2{x}^2=a^2{y}^2$: $b^2\left(\frac{a^2}{2}\right)=a^2\left(\frac{a^2}{2}-c^2\right)$. [3, 5]  
18. Dividing by $a^2$ (since $a\ne 0$) gives $\frac{b^2}{2}=\frac{a^2}{2}-c^2$. 
19. Multiplying by $2$ yields $b^2=a^2-2c^2$. 
20. Rearranging this equation gives $a^2-b^2=2c^2$.