If two tangents drawn from a point (α, β) lying on the ellipse 25x² + 4y² = 1 to the parabola y² = 4

Q: Why can't we just use the discriminant of the quadratic equation to find the tangents?

The discriminant method is typically used to find the condition for a line to be tangent to a curve. Here, we already know the lines are tangents. We are using the point-tangent relationship $y = mx + \frac{1}{m}$ to find the slopes $m$ that allow the tangent to pass through a specific point $(\alpha, \beta)$. The quadratic equation in $m$ arises naturally from this condition.

Q: How do I know which root of the quadratic for $\alpha$ to reject?

Always check the geometric constraints! The point $(\alpha, \beta)$ lies on the ellipse $25x^2 + 4y^2 = 1$. This implies $25\alpha^2 \le 1$, so $|\alpha| \le \frac{1}{5}$. When you solve the quadratic for $\alpha$, one root will fall outside this range, making it physically impossible for that point to be on the ellipse.

Q: Is there a faster way to relate $\alpha$ and $\beta$ without solving for $m$?

While you could use the property of the director circle or other advanced locus properties, the algebraic approach using the sum and product of roots is the most reliable and direct method for this type of JEE problem. It minimizes the chance of missing a subtle geometric detail.

Question

If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25x^2 + 4y^2 = 1$ to the parabola $y^2 = 4x$ are such that the slope of one tangent is four times the other, then the value of $(10\alpha + 5)^2 + (16\beta^2 + 50)^2$ equals ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at two elegant curves: an ellipse defined by $25x^2 + 4y^2 = 1$ and a parabola defined by $y^2 = 4x$.

We are interested in a specific point $(\alpha, \beta)$ that lies on the ellipse. From this point, we draw two tangents to the parabola.

The problem states that one tangent is four times as steep as the other. This condition serves as our primary constraint.

### The Tangent Toolkit

First, we identify the general equation of a tangent to the parabola $y^2 = 4ax$. For our parabola $y^2 = 4x$, we have $a = 1$.

The standard equation for a tangent with slope $m$ is $y = mx + \frac{a}{m}$, which simplifies to:

$$y = mx + \frac{1}{m}$$

Since these tangents must pass through our point $(\alpha, \beta)$, we substitute $x = \alpha$ and $y = \beta$ into the equation:

$$\beta = m\alpha + \frac{1}{m}$$

To clear the fraction, we multiply by $m$, resulting in the quadratic equation:

$$\alpha m^2 - \beta m + 1 = 0$$

This quadratic is the heart of the problem. Its roots, $m_1$ and $m_2$, represent the slopes of the two tangents drawn from $(\alpha, \beta)$.

### The Power of Roots

From the theory of equations, for a quadratic $Am^2 + Bm + C = 0$, the sum of roots is $-\frac{B}{A}$ and the product is $\frac{C}{A}$. Applying this to our equation, we obtain:

$$m_1 + m_2 = \frac{\beta}{\alpha} \quad 	ext{and} \quad m_1 m_2 = \frac{1}{\alpha}$$

The problem provides the condition that one slope is four times the other. Let $m_1 = 4m_2$. Substituting this into our relations, we get:

$$5m_2 = \frac{\beta}{\alpha} \quad 	ext{and} \quad 4m_2^2 = \frac{1}{\alpha}$$

By isolating $m_2$ in the first equation, $m_2 = \frac{\beta}{5\alpha}$, and substituting it into the second, we find:

$$4\left(\frac{\beta}{5\alpha}ight)^2 = \frac{1}{\alpha} \Rightarrow 4\beta^2 = 25\alpha$$

This equation acts as our bridge between the parabola's tangents and the geometry of the ellipse.

### The Final Convergence

We return to the ellipse equation $25\alpha^2 + 4\beta^2 = 1$. Substituting our constraint $4\beta^2 = 25\alpha$, we get:

$$25\alpha^2 + 25\alpha - 1 = 0$$

Solving this quadratic for $\alpha$ using the quadratic formula:

$$\alpha = \frac{-25 \pm \sqrt{625 - 4(25)(-1)}}{50} = \frac{-25 \pm \sqrt{725}}{50} = \frac{-5 \pm \sqrt{29}}{10}$$

We must reject the negative root because the ellipse $25x^2 + 4y^2 = 1$ restricts $x$ to the interval $[-\frac{1}{5}, \frac{1}{5}]$. Thus, we take:

$$\alpha = \frac{-5 + \sqrt{29}}{10}$$

### Final Calculation

We evaluate the expression $(10\alpha + 5)^2 + (16\beta^2 + 50)^2$. Since $10\alpha = -5 + \sqrt{29}$, it follows that $10\alpha + 5 = \sqrt{29}$, and:

$$(10\alpha + 5)^2 = 29$$

For the second part, $16\beta^2 = 4(4\beta^2) = 4(25\alpha) = 100\alpha$. Substituting $\alpha$:

$$100\left(\frac{-5 + \sqrt{29}}{10}ight) = 10(-5 + \sqrt{29}) = -50 + 10\sqrt{29}$$

Thus, $16\beta^2 + 50 = 10\sqrt{29}$, and its square is:

$$(10\sqrt{29})^2 = 100 	imes 29 = 2900$$

Adding these results together, we find the final value:

$$29 + 2900 = \mathbf{2929}$$

Enter Numerical Value:

Visualized Solution

The Curves: Ellipse and Parabola

Point $(α, β)$ on the Ellipse

Tangents to the Parabola

Standard Tangent Equation

Passing Through $(α, β)$

Quadratic Equation in $m$

Roots of the Quadratic

Applying the Given Condition

Finding $m_{2}$

Connecting $α$ and $β$

Substituting into Ellipse Equation

Solving for $α$

Choosing the Correct $α$

Calculating $(10 α + 5)^{2}$

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Toolkit

The Power of Roots

The Final Convergence

Final Calculation

Similar Questions

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If $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}, m_{1} \neq = m_{2}$ are two common tangents of circle $x^{2} + y^{2} = 2$ and parabola $y^{2} = x$ , then the value of $8∣ m_{1} m_{2} ∣$ is equal to

Enter Numerical Value:

Visualized Solution

The Curves: Ellipse and Parabola

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (α,β) on the Ellipse

Tangents to the Parabola

Standard Tangent Equation

Passing Through .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (α,β)

Quadratic Equation in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Roots of the Quadratic

Applying the Given Condition

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2​

Connecting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Substituting into Ellipse Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Choosing the Correct .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (10α+5)2

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Toolkit

The Power of Roots

The Final Convergence

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=m1​x+c1​ and y=m2​x+c2​,m1​=m2​ are two common tangents of circle x2+y2=2 and parabola y2=x, then the value of 8∣m1​m2​∣ is equal to

Point $(α, β)$ on the Ellipse

Passing Through $(α, β)$

Quadratic Equation in $m$

Finding $m_{2}$

Connecting $α$ and $β$

Solving for $α$

Choosing the Correct $α$

Calculating $(10 α + 5)^{2}$

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