NIMCET 2024 — Mathematics PYQ

Step 1: Express $x^2$ from the Hyperbola equation

$$\frac{y^2}{9}-\frac{x^2}{16}=1⟹\frac{x^2}{16}=\frac{y^2}{9}-1$$

$$x^2=16\left(\frac{y^2-9}{9}\right)=\frac{16}{9}(y^2-9)$$

Step 2: Substitute $x^2$ into the Ellipse equation

$$\frac{\frac{16}{9}(y^2-9)}{4}+\frac{(y-4{)}^2}{16}=1$$

$$\frac{4}{9}(y^2-9)+\frac{y^2-8y+16}{16}=1$$

Multiply the entire equation by 144 (LCM of 9 and 16):

$$64(y^2-9)+9(y^2-8y+16)=144$$

$$64y^2-576+9y^2-72y+144=144$$

$$73y^2-72y-576=0$$

Step 3: Check the validity of $y$

For a real intersection point, we must have $x^2\ge 0$.

From Step 1: $x^2=\frac{16}{9}(y^2-9)\ge 0⟹y^2\ge 9⟹y\le -3\text{ or }y\ge 3$.

Solving $73y^2-72y-576=0$ using the quadratic formula:

$$y=\frac{72\pm \sqrt{(-72{)}^2-4(73)(-576)}}{2(73)}$$

$$y=\frac{72\pm \sqrt{5184+173376}}{146}=\frac{72\pm \sqrt{178560}}{146}$$

$$y\approx \frac{72\pm 422.56}{146}$$

• Case 1: $y_1\approx \frac{494.56}{146}\approx 3.38$ (Valid, as $3.38\ge 3$)

• Case 2: $y_2\approx \frac{-350.56}{146}\approx -2.40$ (Invalid, as it does not satisfy $y\le -3$)

Step 4: Count the intersection points

For the single valid value of $y$ ($y\approx 3.38$):

x^2 = \frac{16}{9}((3.38)^2 - 9) > 0

This gives exactly 2 real values of $x$ ($x=\pm \sqrt{\text{positive value}}$).

Final Answer

The curves intersect at exactly 2 points. The correct option is C.