Let P(a, b) be a point on the parabola y² = 8x such that the tangent at P passes through the centre

Q: Why can't we just use the slope-intercept form $y = mx + c$ for the tangent?

You absolutely could, but it is a trap! Using $y = mx + c$ introduces the slope $m$ as an unknown, and then you have to relate $c$ to $m$ using the condition of tangency ($c = A/m$). It creates a more complex algebraic path. Parametric form $t$ encapsulates both the slope and the point in a single variable, making the algebra elegant and direct.

Q: How did we get the center $(5, 7)$ so quickly?

By comparing the given equation $x^2 + y^2 - 10x - 14y + 65 = 0$ with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$. The center is $(-g, -f)$. Since $2g = -10$, $g = -5$, so $-g = 5$. Similarly, $2f = -14$, $f = -7$, so $-f = 7$. It is just a matter of careful coefficient matching!

Q: What if the quadratic equation for $t$ had yielded imaginary roots?

That would imply that no such tangent exists! In geometry, if the math leads to non-real roots, it physically means the line you are trying to construct cannot exist under the given constraints. Always trust the algebra—it is the ultimate arbiter of geometric reality.

Question

Let $P(a, b)$ be a point on the parabola $y^2 = 8x$ such that the tangent at $P$ passes through the centre of the circle $x^2 + y^2 - 10x - 14y + 65 = 0$. Let $A$ be the product of all possible values of $a$ and $B$ be the product of all possible values of $b$. Then the value of $A + B$ is equal to :

Options:

(A) 0

(B) 25

(C) 40

(D) 65

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The parabola is defined by the equation $y^2 = 8x$. We seek a point $P(a, b)$ on this curve such that the tangent at $P$ passes through the center of the circle $x^2 + y^2 - 10x - 14y + 65 = 0$.

### Locating the Heart of the Circle

To find the center of the circle, we compare the given equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$. By matching coefficients, we identify $2g = -10$ and $2f = -14$.

This reveals the center of our circle at $C(-g, -f) = (5, 7)$. This point is our target; our tangent line must pass through it.

### The Elegance of Parametric Coordinates

For the parabola $y^2 = 4Ax$, we identify $4A = 8$, which gives $A = 2$. Any point $P$ on this parabola can be described using the parameter $t$ as:

$$P = (At^2, 2At) = (2t^2, 4t)$$

This substitution reduces our two-dimensional problem into a single-variable equation in terms of $t$.

### The Tangent Bridge

The standard formula for a tangent to $y^2 = 4Ax$ at parameter $t$ is $ty = x + At^2$. Substituting $A = 2$, our tangent equation becomes:

$$ty = x + 2t^2$$

Since this line must pass through the center $C(5, 7)$, we substitute $x = 5$ and $y = 7$ into the equation:

$$7t = 5 + 2t^2$$

Rearranging this yields the quadratic equation:

$$2t^2 - 7t + 5 = 0$$

### The Quadratic Climax

Factoring the quadratic equation $2t^2 - 7t + 5 = 0$ gives:

$$(2t - 5)(t - 1) = 0$$

This results in two distinct values for our parameter: $t = 1$ and $t = \frac{5}{2}$. These values correspond to the two points on the parabola where the tangent line passes through the center of the circle.

### The Final Tally

For $t = 1$, the point is $(2(1)^2, 4(1)) = (2, 4)$. For $t = \frac{5}{2}$, the point is:

$$P = \left(2\left(\frac{5}{2}ight)^2, 4\left(\frac{5}{2}ight)ight) = (12.5, 10)$$

The product of the $x$-coordinates is $2 	imes 12.5 = 25$. The product of the $y$-coordinates is $4 	imes 10 = 40$.

Summing these values, we find the final result:

**$25 + 40 = 65$**

Select Answer:

Visualized Solution

Visualizing the Geometry

Finding the Circle's Center

Parametric Form of Parabola

Equation of the Tangent

Substituting the Center Point

Solving for the Parameter $t$

Calculating Product $A$

Calculating Product $B$

Final Result: $A + B$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Locating the Heart of the Circle

The Elegance of Parametric Coordinates

The Tangent Bridge

The Quadratic Climax

The Final Tally

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