JEE Mathematics 2023 Question with Solution

Given: The parabola is

$$y^2 = 2ax$$

and the tangent is drawn at $P(b,c)$ where $b,c \in \mathbb{N}$.

Find: The value of

$$\sum_{a \in S} a$$

when the area of the triangle formed by the tangent and the lines $x=b$ and $y=0$ is $16$ square units.

Since $P(b,c)$ lies on the parabola,

$$c^2 = 2ab \qquad \text{(1)}$$

The tangent to $y^2=2ax$ at $(x_1,y_1)=(b,c)$ is

$$yy_1 = 2a\left(\frac{x+x_1}{2}\right)$$

So,

$$yc = a(x+b)$$

To find the point where the tangent meets the $x$-axis, put $y=0$:

$$0 = a(x+b)$$

Hence,

$$x=-b$$

So the tangent meets the $x$-axis at $B(-b,0)$.

Also, the line $x=b$ meets the $x$-axis at $A(b,0)$. Therefore,

$$AB = b-(-b)=2b$$

and

$$AP = c$$

Now area of triangle $PBA$ is

$$\text{Area} = \frac{1}{2} \times AB \times AP$$

Given area is $16$, so

$$\frac{1}{2} \times 2b \times c = 16$$

which gives

$$bc=16$$

Since $b,c \in \mathbb{N}$, the possible pairs are

$$(1,16),\ (2,8),\ (4,4),\ (8,2),\ (16,1)$$

From equation $\text{(1)}$,

$$a = \frac{c^2}{2b}$$

Now check which pairs give $a \in \mathbb{N}$:

$$(1,16) \Rightarrow a=\frac{16^2}{2\cdot 1}=128$$ $$(2,8) \Rightarrow a=\frac{8^2}{2\cdot 2}=16$$ $$(4,4) \Rightarrow a=\frac{4^2}{2\cdot 4}=2$$ $$(8,2) \Rightarrow a=\frac{2^2}{2\cdot 8}=\frac{1}{4}$$ $$(16,1) \Rightarrow a=\frac{1^2}{2\cdot 16}=\frac{1}{32}$$

So the valid values of $a$ are $128, 16, 2$.

Therefore,

$$\sum_{a \in S} a = 128+16+2 = 146$$

So, the required answer is $146$.

Given: Tangent at $P(b,c)$ to the parabola

$$y^2=2ax$$

forms a triangle with the lines $x=b$ and $y=0$ having area $16$.

Find: Sum of all natural-number values of $a$.

For the parabola $y^2=2ax$, substituting the point $P(b,c)$ gives

$$c^2=2ab$$

The tangent at $P(b,c)$ is

$$yc=a(x+b)$$

At the $x$-axis, $y=0$, so the tangent cuts the axis at $x=-b$. Thus the base of the triangle between $x=-b$ and $x=b$ is $2b$, and the height from $P(b,c)$ to the $x$-axis is $c$.

Hence,

$$\text{Area} = \frac{1}{2}\times 2b \times c = bc = 16$$

So we need positive integer factor pairs of $16$. Using

$$a=\frac{c^2}{2b}$$

only the pairs $(1,16), (2,8), (4,4)$ make $a$ a natural number, giving

$$a=128,\ 16,\ 2$$

Therefore,

$$128+16+2=146$$

Thus, the answer is $146$.