Animated Solution for Mathematics - Differentiation: Let the normal at a point $P$ on the curve $y^2-3x^2+y+10=0$ intersect the $y$-axis at $\left(0,\frac{3}{2}\right)$. If $m$ is the slope of the tangent at $P$ to the curve, then $|m|$ is equal to

A Journey into Calculus

Welcome, fellow explorers of mathematics! Today, we are going to peel back the layers of a beautiful problem.

Imagine you are standing before the curve defined by the equation $y^2-3x^2+y+10=0$. It is not just a collection of symbols; it is a path in the coordinate plane.

We are interested in a specific point $P(x_1,y_1)$ on this path. At this point, a tangent line kisses the curve, and a normal line stands tall, perpendicular to that tangent. Our goal is to find the absolute value of the slope of that tangent, $|m|$.

The Calculus Toolkit

To find the slope of the tangent, $m$, we need the derivative $\frac{dy}{dx}$. Since our curve is given implicitly, we do not need to isolate $y$.

Instead, we use the power of implicit differentiation. Let us differentiate the entire equation $y^2-3x^2+y+10=0$ with respect to $x$.

Applying the chain rule, the derivative of $y^2$ is $2y\frac{dy}{dx}$, the derivative of $-3x^2$ is $-6x$, and the derivative of $y$ is simply $\frac{dy}{dx}$. The derivative of the constant $10$ is, of course, zero.

Thus, we get:

By grouping the terms with $\frac{dy}{dx}$, we find that $(2y+1)\frac{dy}{dx}=6x$. Therefore, the slope of the tangent at any point $(x,y)$ is given by:

The Geometric Bridge

Now, let us consider the normal line. We know that the normal is perpendicular to the tangent.

If the tangent has a slope $m$, the normal must have a slope $m_n=-\frac{1}{m}$. Substituting our expression for $m$, we get:

With this slope and the point $P(x_1,y_1)$, we can write the equation of the normal line using the point-slope form: $y-y_1=m_n(x-x_1)$. Substituting our $m_n$, the equation becomes:

The Intersection

We are told that this normal line intersects the $y$-axis at the point $(0,\frac{3}{2})$. This is the key that unlocks the door!

Since this point lies on the normal line, it must satisfy the equation. Let us substitute $x=0$ and $y=\frac{3}{2}$ into our normal equation:

Notice the beauty of the algebra here: the $-x_1$ on the right side cancels with the $x_1$ in the denominator, and the negative signs cancel out. We are left with:

Multiplying the entire equation by $6$ gives us $9-6y_1=2y_1+1$. Solving for $y_1$, we find $8y_1=8$, which means $y_1=1$.

The Final Reveal

Now that we have $y_1=1$, we return to the original curve equation to find $x_1$. Substituting $y_1=1$ into $y^2-3x^2+y+10=0$, we get:

This simplifies to $12-3x_1^2=0$. This leads to $3x_1^2=12$, or $x_1^2=4$. Thus, $x_1=\pm 2$.

Finally, we calculate the slope $m$ at $P(x_1,1)$. Using $m=\frac{6x_1}{2y_1+1}$: If $x_1=2$, then $m=\frac{6(2)}{2(1)+1}=\frac{12}{3}=4$. If $x_1=-2$, then $m=\frac{6(-2)}{2(1)+1}=-4$.

In both cases, the absolute value $|m|$ is 4. We have arrived at our destination!