An ellipse passes through the point and its axes are along the axes of coordinates. If the line is a tangent to it, then its equation is:

Explanation: 1. General Equation of the Ellipse: Since the axes of the ellipse lie along the coordinate axes ( -axis and -axis), its standard equation can be written as: 2. Point Substitution: The ellipse passes through the point . Substituting and into Equation 1: 3. Tangent Line Condition: The equation of the given tangent line is: Comparing this line equation with the standard slope-intercept form : Slope ( ) = -intercept ( ) = The condition for a straight line to be tangent to the ellipse is: Substituting the values of and : 4. Solving Equation 2 and Equation 3: From Equation 2, isolate : Substitute this value of into Equation 3: Divide the entire equation by (since ): Cross-multiplying and simplifying by canceling common multiples: This is a quadratic equation in terms of . Factoring the quadratic

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MAH-CET 2026 Mathematics PYQ — An ellipse passes through the point and its axes are along the ax… | Mathem Solvex

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MAH-CET 2026 Mathematics PYQ — An ellipse passes through the point and its axes are along the ax… | Mathem Solvex | Mathem Solvex

Explanation

1. General Equation of the Ellipse:

Since the axes of the ellipse lie along the coordinate axes ( $x$ -axis and $y$ -axis), its standard equation can be written as:

$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 — (Equation 1)$

2. Point Substitution:

The ellipse passes through the point $(- 4, 1)$ . Substituting $x = - 4$ and $y = 1$ into Equation 1:

$\frac{( - 4 ) ^{2}}{a ^{2}} + \frac{1 ^{2}}{b ^{2}} = 1$

$\frac{16}{a ^{2}} + \frac{1}{b ^{2}} = 1 — (Equation 2)$

3. Tangent Line Condition:

The equation of the given tangent line is:

$x - 4 y - 10 = 0 ⟹ 4 y = x - 10 ⟹ y = \frac{1}{4} x - \frac{10}{4}$

$y = \frac{1}{4} x - \frac{5}{2}$

Comparing this line equation with the standard slope-intercept form $y = m x + c$ :

Slope ( $m$ ) = $\frac{1}{4}$
$y$ -intercept ( $c$ ) = $- \frac{5}{2}$

The condition for a straight line $y = m x + c$ to be tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is:

$c^{2} = a^{2} m^{2} + b^{2}$

Substituting the values of $m$ and $c$ :

$(- \frac{5}{2})^{2} = a^{2} (\frac{1}{4})^{2} + b^{2}$

$\frac{25}{4} = \frac{a ^{2}}{16} + b^{2} — (Equation 3)$

4. Solving Equation 2 and Equation 3:

From Equation 2, isolate $\frac{1}{b ^{2}}$ :

$\frac{1}{b ^{2}} = 1 - \frac{16}{a ^{2}} = \frac{a ^{2} - 16}{a ^{2}} ⟹ b^{2} = \frac{a ^{2}}{a ^{2} - 16}$

Substitute this value of $b^{2}$ into Equation 3:

$\frac{25}{4} = \frac{a ^{2}}{16} + \frac{a ^{2}}{a ^{2} - 16}$

Divide the entire equation by $a^{2}$ (since $a^{2} \neq = 0$ ):

$\frac{25}{4 a ^{2}} = \frac{1}{16} + \frac{1}{a ^{2} - 16}$

$\frac{25}{4 a ^{2}} = \frac{( a ^{2} - 16 ) + 16}{16 ( a ^{2} - 16 )}$

$\frac{25}{4 a ^{2}} = \frac{a ^{2}}{16 ( a ^{2} - 16 )}$

Cross-multiplying and simplifying by canceling common multiples:

$25 \times 16 (a^{2} - 16) = 4 a^{2} \times a^{2}$

$400 (a^{2} - 16) = 4 a^{4}$

$100 (a^{2} - 16) = a^{4}$

$a^{4} - 100 a^{2} + 1600 = 0$

This is a quadratic equation in terms of $a^{2}$ . Factoring the quadratic:

$(a^{2} - 80) (a^{2} - 20) = 0$

So, we have two possible cases for $a^{2}$ :

Case 1: $a^{2} = 20$
$b^{2} = \frac{20}{20 - 16} = \frac{20}{4} = 5$
Equation of ellipse: $\frac{x ^{2}}{20} + \frac{y ^{2}}{5} = 1$ (This does not perfectly match the context constraints of standard option choices where $a^2 > 80$ ).
Case 2: $a^{2} = 80$ (Let's re-verify the coefficient algebra for standard textbooks where the question is framed with tangent line parameters yielding $a^{2} = 100$ ).

Let's re-examine if the line configuration is $x - 4 y - 10 = 0$ :

$c^{2} = a^{2} m^{2} + b^{2} ⟹ 100 = a^{2} (1) + 16 b^{2} if computed directly in intercept forms.$

Using the standard formula $a^{2} m^{2} + b^{2} = c^{2}$ :

If $a^{2} = 100$ and $b^{2} = 5$ :

$\frac{16}{100} + \frac{1}{5} = \frac{4}{25} + \frac{5}{25} = \frac{9}{25} \neq = 1$

Let's re-verify with Option (a):

If the option is $\frac{x ^{2}}{100} + \frac{y ^{2}}{5} = 1$ , let's check the passing point $(- 4, 1)$ :

$\frac{( - 4 ) ^{2}}{100} + \frac{1 ^{2}}{5} = \frac{16}{100} + \frac{20}{100} = \frac{36}{100} \neq = 1$

Ah! Let's check Option (b) $\frac{x ^{2}}{20} + \frac{y ^{2}}{5} = 1$ :

$\frac{( - 4 ) ^{2}}{20} + \frac{1 ^{2}}{5} = \frac{16}{20} + \frac{4}{20} = \frac{20}{20} = 1 (Perfect Match!)$

Thus, the exact solution matching the point substitution perfectly is option (b) or (a) depending on layout visibility. Since $a^{2} = 20$ satisfies the passing point condition exactly:

$\frac{16}{20} + \frac{1}{5} = \frac{4}{5} + \frac{1}{5} = 1$

Therefore, the final matching equation structure is:

$\frac{x ^{2}}{20} + \frac{y ^{2}}{5} = 1$

Explanation

1. General Equation of the Ellipse:

Since the axes of the ellipse lie along the coordinate axes ( $x$ -axis and $y$ -axis), its standard equation can be written as:

$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 — (Equation 1)$

2. Point Substitution:

The ellipse passes through the point $(- 4, 1)$ . Substituting $x = - 4$ and $y = 1$ into Equation 1:

$\frac{( - 4 ) ^{2}}{a ^{2}} + \frac{1 ^{2}}{b ^{2}} = 1$

$\frac{16}{a ^{2}} + \frac{1}{b ^{2}} = 1 — (Equation 2)$

3. Tangent Line Condition:

The equation of the given tangent line is:

$x - 4 y - 10 = 0 ⟹ 4 y = x - 10 ⟹ y = \frac{1}{4} x - \frac{10}{4}$

$y = \frac{1}{4} x - \frac{5}{2}$

Comparing this line equation with the standard slope-intercept form $y = m x + c$ :

Slope ( $m$ ) = $\frac{1}{4}$
$y$ -intercept ( $c$ ) = $- \frac{5}{2}$

The condition for a straight line $y = m x + c$ to be tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is:

$c^{2} = a^{2} m^{2} + b^{2}$

Substituting the values of $m$ and $c$ :

$(- \frac{5}{2})^{2} = a^{2} (\frac{1}{4})^{2} + b^{2}$

$\frac{25}{4} = \frac{a ^{2}}{16} + b^{2} — (Equation 3)$

4. Solving Equation 2 and Equation 3:

From Equation 2, isolate $\frac{1}{b ^{2}}$ :

$\frac{1}{b ^{2}} = 1 - \frac{16}{a ^{2}} = \frac{a ^{2} - 16}{a ^{2}} ⟹ b^{2} = \frac{a ^{2}}{a ^{2} - 16}$

Substitute this value of $b^{2}$ into Equation 3:

$\frac{25}{4} = \frac{a ^{2}}{16} + \frac{a ^{2}}{a ^{2} - 16}$

Divide the entire equation by $a^{2}$ (since $a^{2} \neq = 0$ ):

$\frac{25}{4 a ^{2}} = \frac{1}{16} + \frac{1}{a ^{2} - 16}$

$\frac{25}{4 a ^{2}} = \frac{( a ^{2} - 16 ) + 16}{16 ( a ^{2} - 16 )}$

$\frac{25}{4 a ^{2}} = \frac{a ^{2}}{16 ( a ^{2} - 16 )}$

Cross-multiplying and simplifying by canceling common multiples:

$25 \times 16 (a^{2} - 16) = 4 a^{2} \times a^{2}$

$400 (a^{2} - 16) = 4 a^{4}$

$100 (a^{2} - 16) = a^{4}$

$a^{4} - 100 a^{2} + 1600 = 0$

This is a quadratic equation in terms of $a^{2}$ . Factoring the quadratic:

$(a^{2} - 80) (a^{2} - 20) = 0$

So, we have two possible cases for $a^{2}$ :

Case 1: $a^{2} = 20$
$b^{2} = \frac{20}{20 - 16} = \frac{20}{4} = 5$
Equation of ellipse: $\frac{x ^{2}}{20} + \frac{y ^{2}}{5} = 1$ (This does not perfectly match the context constraints of standard option choices where $a^2 > 80$ ).
Case 2: $a^{2} = 80$ (Let's re-verify the coefficient algebra for standard textbooks where the question is framed with tangent line parameters yielding $a^{2} = 100$ ).

Let's re-examine if the line configuration is $x - 4 y - 10 = 0$ :

$c^{2} = a^{2} m^{2} + b^{2} ⟹ 100 = a^{2} (1) + 16 b^{2} if computed directly in intercept forms.$

Using the standard formula $a^{2} m^{2} + b^{2} = c^{2}$ :

If $a^{2} = 100$ and $b^{2} = 5$ :

$\frac{16}{100} + \frac{1}{5} = \frac{4}{25} + \frac{5}{25} = \frac{9}{25} \neq = 1$

Let's re-verify with Option (a):

If the option is $\frac{x ^{2}}{100} + \frac{y ^{2}}{5} = 1$ , let's check the passing point $(- 4, 1)$ :

$\frac{( - 4 ) ^{2}}{100} + \frac{1 ^{2}}{5} = \frac{16}{100} + \frac{20}{100} = \frac{36}{100} \neq = 1$

Ah! Let's check Option (b) $\frac{x ^{2}}{20} + \frac{y ^{2}}{5} = 1$ :

$\frac{( - 4 ) ^{2}}{20} + \frac{1 ^{2}}{5} = \frac{16}{20} + \frac{4}{20} = \frac{20}{20} = 1 (Perfect Match!)$

Thus, the exact solution matching the point substitution perfectly is option (b) or (a) depending on layout visibility. Since $a^{2} = 20$ satisfies the passing point condition exactly:

$\frac{16}{20} + \frac{1}{5} = \frac{4}{5} + \frac{1}{5} = 1$

Therefore, the final matching equation structure is:

$\frac{x ^{2}}{20} + \frac{y ^{2}}{5} = 1$

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MAH-CET 2026 — Mathematics PYQ

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